Lesson Explainer: Indefinite Integrals: The Power Rule Mathematics • 12th Grade

In this explainer, we will learn how to find the indefinite integrals of polynomials and general power functions using the power rule for integration.

Recall that an antiderivative, also known as an inverse derivative or primitive, of a function ๐‘“ is another function ๐น whose derivative is equal to the original function ๐‘“.

Definition: An Antiderivative of a Function

For any function ๐‘“ defined on a subset ๐‘ˆโŠ†โ„ and a differentiable function ๐นโˆถ๐‘ˆโ†’โ„, if we have ๐นโ€ฒ(๐‘ฅ)=๐‘“(๐‘ฅ), then we say that ๐น(๐‘ฅ) is an antiderivative of ๐‘“(๐‘ฅ).

The antiderivative of a function is equivalent to the indefinite integral, which we define as follows.

Definition: The Indefinite Integral

The indefinite integral of ๐‘“(๐‘ฅ) with respect to ๐‘ฅ can be written in terms of an antiderivative ๐น(๐‘ฅ) as ๏„ธ๐‘“(๐‘ฅ)๐‘ฅ=๐น(๐‘ฅ)+,dC where C is also called the constant of integration.

Antiderivatives of ๐‘“ always exist when ๐‘“ is continuous, and there are infinitely many antiderivatives for ๐‘“, obtained by adding the arbitrary constant C to ๐น. This constant, also known as the constant of integration, is very important, as it produces a family of antiderivatives paramaterized by C. In other words, ๐น(๐‘ฅ)+C is the most general function that has a derivative ๐‘“(๐‘ฅ), for all Cโˆˆโ„. For example, the derivative of ๐‘ฅ is given by (๐‘ฅ)โ€ฒ=1.

Thus, we can say that ๐‘ฅ is an antiderivative of 1, but ๐‘ฅ+C is the most general antiderivative of 1, which means that ๐‘ฅ+1, ๐‘ฅ+7, ๐‘ฅ+โˆš2, ๐‘ฅ+๐œ‹, and so on are all also antiderivatives of 1. This is what we call the indefinite integral and is expressed as ๏„ธ1๐‘ฅ=๐‘ฅ+.dC

Similarly, the derivative of ๐‘ฅ๏Šจ is ๏€น๐‘ฅ๏…=2๐‘ฅ,๏Šจ๏Ž˜ which implies that the indefinite integral of 2๐‘ฅ is ๏„ธ2๐‘ฅ๐‘ฅ=๐‘ฅ+.dC๏Šจ

Recall that the derivative satisfies the property (๐‘Ž๐น(๐‘ฅ))โ€ฒ=๐‘Ž๐นโ€ฒ(๐‘ฅ).

This means we can always take out a constant multiple outside the derivative. Thus, ๐‘Ž๐น(๐‘ฅ) is an antiderivative of ๐‘Ž๐‘“(๐‘ฅ), or ๐‘Ž๐น(๐‘ฅ)+C is the general antiderivative of ๐‘Ž๐‘“(๐‘ฅ), for all Cโˆˆโ„, which is the same as the indefinite integral. In other words, if the derivative gets multiplied by a constant, the antiderivative also gets multiplied by the same constant and vice versa. This implies the following property of indefinite integrals: ๏„ธ(๐‘Ž๐‘“(๐‘ฅ))๐‘ฅ=๐‘Ž๏„ธ๐‘“(๐‘ฅ)๐‘ฅ.dd

Using the indefinite integral of 2๐‘ฅ as written above, we can take out the factor of 2 using this property and divide both sides of the expression by 2 to obtain ๏„ธ2๐‘ฅ๐‘ฅ=2๏„ธ๐‘ฅ๐‘ฅ=๐‘ฅ+๏„ธ๐‘ฅ๐‘ฅ=๐‘ฅ2+,ddCdC๏Šจ๏Šจ where we note that we have kept the constant C the same since this is arbitrary and C2 is just another constant. In this explainer, we will be particularly interested in determining indefinite integrals of the form ๏„ธ๐‘ฅ๐‘ฅ๏Šd using the power rule of integration. We can determine this rule directly from the power rule of differentiation. Suppose ๐น(๐‘ฅ)=๐‘ฅ๏Œ, for ๐‘โˆˆโ„. The derivative of this function can be found by the power rule for differentiation as follows: (๐‘ฅ)=๐‘๐‘ฅ.๏Œ๏Ž˜๏Œ๏Šฑ๏Šง

It will be useful to rewrite this as 1๐‘(๐‘ฅ)=1๐‘๏€น๐‘๐‘ฅ๏…๏€ฝ๐‘ฅ๐‘๏‰=๐‘ฅ,๏Œ๏Ž˜๏Œ๏Šฑ๏Šง๏Œ๏Ž˜๏Œ๏Šฑ๏Šง where we have divided by the constant ๐‘โ‰ 0, as a constant multiple of a function does not affect the derivative or antiderivative. But what if we want to work in reverse? (i.e., given ๐‘ฅ๏Œบ, we want to determine the antiderivative). This means we want to find the most general function that differentiates to give ๐‘ฅ๏Œบ.

We have already shown that the derivative of ๐‘ฅ๐‘๏Œ is ๐‘ฅ๏Œ๏Šฑ๏Šง, for ๐‘โ‰ 0. If we let ๐‘=๐‘›+1, then we have ๏€พ๐‘ฅ๐‘›+1๏Š=๐‘ฅ,๐‘›โ‰ โˆ’1.๏Š๏Šฐ๏Šง๏Ž˜๏Š

Thus, ๐‘ฅ๐‘›+1๏Š๏Šฐ๏Šง is an antiderivative of ๐‘ฅ๏Š, provided ๐‘›โ‰ โˆ’1. We can express this in terms of an indefinite integral in the following definition.

Rule: The Power Rule for Integration

The power rule for integration allows us to determine the indefinite integral of ๐‘ฅ๏Š, provided ๐‘›โ‰ โˆ’1, as follows: ๏„ธ๐‘ฅ๐‘ฅ=๐‘ฅ๐‘›+1+.๏Š๏Š๏Šฐ๏ŠงdC

For example, using this power rule, we can determine the indefinite integral of ๐‘ฅ๏Šจ as follows: ๏„ธ๐‘ฅ๐‘ฅ=๐‘ฅ2+1+=๐‘ฅ3+,๏Šจ๏Šจ๏Šฐ๏Šง๏ŠฉdCC which can be verified directly by differentiating the right-hand side to obtain the integrand ๐‘ฅ๏Šจ.

In the first example, we will determine the indefinite integral of a function involving a positive integer power of ๐‘ฅ using the power rule along with the property that allows us to take a constant multiple outside the integral.

Example 1: The Power Rule of Integration

Determine ๏„ธโˆ’๐‘ฅ๐‘ฅ๏Šฏd.

Answer

In this example, we will determine the indefinite integral of a positive integer power of ๐‘ฅ, in particular the function โˆ’๐‘ฅ๏Šฏ.

In order to determine the integral, we will make use of the following property of indefinite integrals: ๏„ธ(๐‘Ž๐‘“(๐‘ฅ))๐‘ฅ=๐‘Ž๏„ธ๐‘“(๐‘ฅ)๐‘ฅ.dd

We will also make use of the power rule: ๏„ธ๐‘ฅ๐‘ฅ=๐‘ฅ๐‘›+1+,๐‘›โ‰ โˆ’1.๏Š๏Š๏Šฐ๏ŠงdC

We can use the property to take out the factor of (โˆ’1) outside the integral and determine the indefinite integral of ๐‘ฅ๏Šฏ using the power rule: ๏„ธโˆ’๐‘ฅ๐‘ฅ=โˆ’๏„ธ๐‘ฅ๐‘ฅ=โˆ’๏€พ๐‘ฅ9+1๏Š+=โˆ’๐‘ฅ10+.๏Šฏ๏Šฏ๏Šฏ๏Šฐ๏Šง๏Šง๏ŠฆddCC

Now, letโ€™s consider an example where we will determine the indefinite integral of a function involving a negative integer power of ๐‘ฅ using the power rule along with the property that allows us to take a constant multiple outside the integral.

Example 2: Finding the Integration of a Function Using the Power Rule for Integration with a Negative Exponent

Determine ๏„ธโˆ’27๐‘ฅ๐‘ฅ๏Šฑ๏Šฏd.

Answer

In this example, we will determine the indefinite integral of a negative integer power of ๐‘ฅ, in particular the function โˆ’27๐‘ฅ๏Šฑ๏Šฏ.

In order to determine the integral, we will make use of the following property of indefinite integrals: ๏„ธ(๐‘Ž๐‘“(๐‘ฅ))๐‘ฅ=๐‘Ž๏„ธ๐‘“(๐‘ฅ)๐‘ฅ.dd

We will also make use of the power rule: ๏„ธ๐‘ฅ๐‘ฅ=๐‘ฅ๐‘›+1+,๐‘›โ‰ โˆ’1.๏Š๏Š๏Šฐ๏ŠงdC

We can use the property to take the factor of โˆ’27 outside the integral and determine the indefinite integral of ๐‘ฅ๏Šฑ๏Šฏ using the power rule: ๏„ธโˆ’27๐‘ฅ๐‘ฅ=โˆ’27๏„ธ๐‘ฅ๐‘ฅ=โˆ’27๏€พ๐‘ฅโˆ’9+1๏Š+=โˆ’27๏€พ๐‘ฅโˆ’8๏Š+=2๐‘ฅ56+=128๐‘ฅ+.๏Šฑ๏Šฏ๏Šฑ๏Šฏ๏Šฑ๏Šฏ๏Šฐ๏Šง๏Šฑ๏Šฎ๏Šฑ๏Šฎ๏Šฑ๏ŠฎddCCCC

This result is valid for all ๐‘ฅโ‰ 0, since we require the integrand and integral to be continuous and well defined.

We can use the power rule to determine the indefinite integral of any power of ๐‘ฅ and not just integers, provided that power is not equal to โˆ’1. In the next example, we will determine the indefinite integral of a function involving a positive fractional power of ๐‘ฅ by rewriting the radical in terms of a power of ๐‘ฅ and using the power rule along with the property that allows us to take a constant multiple outside the integral.

Example 3: Finding the General Antiderivative of a Function Using the Power Rule of Integration with Fractional Exponents

Determine ๏„ธ7โˆš๐‘ฅ๐‘ฅ๏Šฉd.

Answer

In this example, we will determine the indefinite integral of a positive fractional power of ๐‘ฅ, in particular the function 7โˆš๐‘ฅ๏Šฉ.

Letโ€™s first rewrite the integrand, by noting that โˆš๐‘ฅ=๐‘ฅ๏Ž ๏Žก, as 7โˆš๐‘ฅ=7๏€น๐‘ฅ๏…=7๐‘ฅ.๏Šฉ๏Šฉ๏Ž ๏Žก๏Žข๏Žก

In order to determine the integral, we will make use of the following property of indefinite integrals: ๏„ธ(๐‘Ž๐‘“(๐‘ฅ))๐‘ฅ=๐‘Ž๏„ธ๐‘“(๐‘ฅ)๐‘ฅ.dd

We will also make use of the power rule: ๏„ธ๐‘ฅ๐‘ฅ=๐‘ฅ๐‘›+1+,๐‘›โ‰ โˆ’1.๏Š๏Š๏Šฐ๏ŠงdC

We can use the property to take the factor of 7 outside the integral and determine the indefinite integral of ๐‘ฅ๏Žข๏Žก using the power rule. Following this, we can rewrite the final answer back in terms of the square root: ๏„ธ7โˆš๐‘ฅ๐‘ฅ=๏„ธ7๐‘ฅ๐‘ฅ=7๏„ธ๐‘ฅ๐‘ฅ=7๏‚๐‘ฅ+1๏Ž+=7๏‚๐‘ฅ๏Ž+=7๏2๐‘ฅ5๏+=145๐‘ฅ+=145โˆš๐‘ฅ+.๏Šฉ๏Šฐ๏Šง๏Šฉ๏Šจ๏Šซ๏Šจ๏ŠซdddCCCCC๏Žข๏Žก๏Žข๏Žก๏Žข๏Žก๏Žค๏Žก๏Žค๏Žก๏Žค๏Žก

This result is valid for all ๐‘ฅโ‰ฅ0, since we require the integrand and integral to be continuous and well defined, and the square root is only defined for nonnegative numbers.

Now, letโ€™s consider an example where we will determine the indefinite integral of a function involving a negative fractional power of ๐‘ฅ by rewriting the radical in terms of a power of ๐‘ฅ and by using the power rule along with the property that allows us to take a constant multiple outside the integral.

Example 4: Finding the Integration of a Function Using the Power Rule with Roots

Determine ๏„ธ6โˆš๐‘ฅ๐‘ฅ๏Žงd.

Answer

In this example, we will determine the indefinite integral of a negative fractional power of ๐‘ฅ, in particular the function 6โˆš๐‘ฅ๏Žง.

First, letโ€™s rewrite the integrand, by noting that ๏Žง๏Ž ๏Žงโˆš๐‘ฅ=๐‘ฅ, as 6โˆš๐‘ฅ=6๐‘ฅ=6๐‘ฅ.๏Žง๏Ž ๏Žง๏Žช๏Ž ๏Žง

In order to determine the integral, we will make use of the following property of indefinite integrals: ๏„ธ(๐‘Ž๐‘“(๐‘ฅ))๐‘ฅ=๐‘Ž๏„ธ๐‘“(๐‘ฅ)๐‘ฅ.dd

We will also make use of the power rule: ๏„ธ๐‘ฅ๐‘ฅ=๐‘ฅ๐‘›+1+,๐‘›โ‰ โˆ’1.๏Š๏Š๏Šฐ๏ŠงdC

We can use the property to take out the factor of 6 outside the integral and determine the indefinite integral of ๐‘ฅ๏Žช๏Ž ๏Žง using the power rule: ๏„ธ6โˆš๐‘ฅ๐‘ฅ=๏„ธ6๐‘ฅ๐‘ฅ=6๏„ธ๐‘ฅ๐‘ฅ=6๏‚๐‘ฅ+1๏Ž+=6๏‚๐‘ฅ๏Ž+=6๏8๐‘ฅ7๏+=487๐‘ฅ+.๏Žง๏Žช๏Ž ๏Žง๏Žช๏Ž ๏Žง๏Žช๏Ž ๏Žง๏Žฆ๏Žง๏Žฆ๏Žง๏Žฆ๏ŽงdddCCCC๏Šฐ๏Šง๏Šฑ๏Šง๏Šฎ๏Šญ๏Šฎ

This result is valid for all ๐‘ฅ>0, since we require the integrand and integral to be continuous and well defined, and the 8th root is only defined for nonnegative numbers.

Recall that the derivative is a linear operation, as it satisfies (๐น(๐‘ฅ)+๐บ(๐‘ฅ))โ€ฒ=๐นโ€ฒ(๐‘ฅ)+๐บโ€ฒ(๐‘ฅ).

This also implies a similar rule for indefinite integrals: ๏„ธ(๐‘“(๐‘ฅ)+๐‘”(๐‘ฅ))๐‘ฅ=๏„ธ๐‘“(๐‘ฅ)๐‘ฅ+๏„ธ๐‘”(๐‘ฅ)๐‘ฅ.ddd

Therefore, in order to determine the indefinite integral of a sum of functions, we just find an indefinite integral of each part separately and add the results together, not forgetting the +C at the end. Usually, we would obtain multiple constants for each part from the process of integration, but we can combine these into one constant. We can also combine this property with the one that allows us to take constants outside the integral.

Property: Linearity Property of Integration

For any continuous functions ๐‘“ and ๐‘” defined on a subset ๐‘ˆโŠ†โ„, we have the linearity property ๏„ธ(๐‘Ž๐‘“(๐‘ฅ)+๐‘๐‘”(๐‘ฅ))๐‘ฅ=๐‘Ž๏„ธ๐‘“(๐‘ฅ)๐‘ฅ+๐‘๏„ธ๐‘”(๐‘ฅ)๐‘ฅ,ddd for ๐‘Ž,๐‘โˆˆโ„.

The power rule for integration along with this linearity property allow us to determine the indefinite integral involving sums of different powers of ๐‘ฅ including polynomial, reciprocal, and radical functions. For example, we can determine the indefinite integral of the linear function 6๐‘ฅ+8 as follows: ๏„ธ(6๐‘ฅ+8)๐‘ฅ=๏„ธ6๐‘ฅ๐‘ฅ+๏„ธ8๐‘ฅ=6๏„ธ๐‘ฅ๐‘ฅ+8๏„ธ1๐‘ฅ=6๏€พ๐‘ฅ1+1๏Š+8๐‘ฅ+=6๏€พ๐‘ฅ2๏Š+8๐‘ฅ+=3๐‘ฅ+8๐‘ฅ+.dddddCCC๏Šง๏Šฐ๏Šง๏Šจ๏Šจ

In the next example, we will determine the indefinite integral of a polynomial function using the properties of linearity and the power rule for integrals.

Example 5: Finding the Integration of a Polynomial Function Using the Power Rule

Determine ๏„ธ๏€น25๐‘ฅโˆ’65๐‘ฅ+36๏…๐‘ฅ๏Šจd.

Answer

In this example, we will determine the indefinite integral of the polynomial function 25๐‘ฅโˆ’65๐‘ฅ+36๏Šจ.

In order to determine the integral, we will make use of the following properties of indefinite integrals: ๏„ธ(๐‘“(๐‘ฅ)+๐‘”(๐‘ฅ))๐‘ฅ=๏„ธ๐‘“(๐‘ฅ)๐‘ฅ+๏„ธ๐‘”(๐‘ฅ)๐‘ฅ,๏„ธ(๐‘Ž๐‘“(๐‘ฅ))๐‘ฅ=๐‘Ž๏„ธ๐‘“(๐‘ฅ)๐‘ฅ.ddddd

We will also make use of the power rule: ๏„ธ๐‘ฅ๐‘ฅ=๐‘ฅ๐‘›+1+,๐‘›โ‰ โˆ’1.๏Š๏Š๏Šฐ๏ŠงdC

Using the first property, we can split the given integral into three parts. We can then use the second property to take the appropriate factors outside the integral and determine the indefinite integral of the different terms using the power rule: ๏„ธ๏€น25๐‘ฅโˆ’65๐‘ฅ+36๏…๐‘ฅ=๏„ธ25๐‘ฅ๐‘ฅ+๏„ธโˆ’65๐‘ฅ๐‘ฅ+๏„ธ36๐‘ฅ=25๏„ธ๐‘ฅ๐‘ฅโˆ’65๏„ธ๐‘ฅ๐‘ฅ+36๏„ธ1๐‘ฅ=25๏€พ๐‘ฅ2+1๏Šโˆ’65๏€พ๐‘ฅ1+1๏Š+36๏€พ๐‘ฅ0+1๏Š=25๏€พ๐‘ฅ3๏Šโˆ’65๏€พ๐‘ฅ2๏Š+36๐‘ฅ=253๐‘ฅโˆ’652๐‘ฅ+36๐‘ฅ+.๏Šจ๏Šจ๏Šจ๏Šจ๏Šฐ๏Šง๏Šง๏Šฐ๏Šง๏Šฆ๏Šฐ๏Šง๏Šฉ๏Šจ๏Šฉ๏ŠจdddddddC

Note that we would obtain a constant of integration for each part from the process of integration, but we can combine these into one constant, C.

Now, letโ€™s consider an example where we will determine the indefinite integral of a polynomial function by distributing two brackets using the properties of linearity and the power rule for integrals.

Example 6: Finding the Integration of a Polynomial Involving Multiplying Two Brackets and Applying the Power Rule

Determine ๏„ธ(๐‘ฅ+4)๏€น๐‘ฅโˆ’4๐‘ฅ+16๏…๐‘ฅ๏Šจd.

Answer

In this example, we will determine the indefinite integral of the polynomial function (๐‘ฅ+4)๏€น๐‘ฅโˆ’4๐‘ฅ+16๏…๏Šจ.

Letโ€™s first simplify the integrand by distributing the brackets: (๐‘ฅ+4)๏€น๐‘ฅโˆ’4๐‘ฅ+16๏…=๐‘ฅโˆ’4๐‘ฅ+16๐‘ฅ+4๐‘ฅโˆ’16๐‘ฅ+64=๐‘ฅ+64.๏Šจ๏Šฉ๏Šจ๏Šจ๏Šฉ

In order to determine the integral, we will make use of the following properties of indefinite integrals: ๏„ธ(๐‘“(๐‘ฅ)+๐‘”(๐‘ฅ))๐‘ฅ=๏„ธ๐‘“(๐‘ฅ)๐‘ฅ+๏„ธ๐‘”(๐‘ฅ)๐‘ฅ,๏„ธ(๐‘Ž๐‘“(๐‘ฅ))๐‘ฅ=๐‘Ž๏„ธ๐‘“(๐‘ฅ)๐‘ฅ.ddddd

We will also make use of the power rule: ๏„ธ๐‘ฅ๐‘ฅ=๐‘ฅ๐‘›+1+,๐‘›โ‰ โˆ’1.๏Š๏Š๏Šฐ๏ŠงdC

Using the first property, we can split the given integral into two parts. We can then use the second property to take the appropriate factors outside the integral and determine the indefinite integral of the different terms using the power rule: ๏„ธ(๐‘ฅ+4)๏€น๐‘ฅโˆ’4๐‘ฅ+16๏…๐‘ฅ=๏„ธ๏€น๐‘ฅ+64๏…๐‘ฅ=๏„ธ๐‘ฅ๐‘ฅ+๏„ธ64๐‘ฅ=๏„ธ๐‘ฅ๐‘ฅ+64๏„ธ1๐‘ฅ=๐‘ฅ3+1+64๐‘ฅ+=๐‘ฅ4+64๐‘ฅ+.๏Šจ๏Šฉ๏Šฉ๏Šฉ๏Šฉ๏Šฐ๏Šง๏ŠชddddddCC

Note that we would obtain a constant of integration for each part from the process of integration, but we can combine these into one constant, C.

In the next example, we will determine the indefinite integral of a rational function using factorization, the properties of linearity, and the power rule for integrals.

Example 7: Finding the Integration of a Rational Function Using the Factorization of the Difference of Two Squares

Determine ๏„ธ๐‘ฅโˆ’225๐‘ฅโˆ’15๐‘ฅ๏Šจd.

Answer

In this example, we will determine the indefinite integral of the rational function ๐‘ฅโˆ’225๐‘ฅโˆ’15๏Šจ.

Letโ€™s first simplify the integrand by noting that the numerator is the difference of two squares and can be written as ๐‘ฅโˆ’225=(๐‘ฅ+15)(๐‘ฅโˆ’15)๏Šจ; thus, the integrand becomes ๐‘ฅโˆ’225๐‘ฅโˆ’15=(๐‘ฅ+15)(๐‘ฅโˆ’15)(๐‘ฅโˆ’15)=๐‘ฅ+15,๏Šจ for ๐‘ฅโ‰ 15. In order to determine the integral, we will make use of the following properties of indefinite integrals: ๏„ธ(๐‘“(๐‘ฅ)+๐‘”(๐‘ฅ))๐‘ฅ=๏„ธ๐‘“(๐‘ฅ)๐‘ฅ+๏„ธ๐‘”(๐‘ฅ)๐‘ฅ,๏„ธ(๐‘Ž๐‘“(๐‘ฅ))๐‘ฅ=๐‘Ž๏„ธ๐‘“(๐‘ฅ)๐‘ฅ.ddddd

We will also make use of the power rule: ๏„ธ๐‘ฅ๐‘ฅ=๐‘ฅ๐‘›+1+,๐‘›โ‰ โˆ’1.๏Š๏Š๏Šฐ๏ŠงdC

Using the first property we can split the given integral into two parts. We can then use the second property to take the appropriate factors outside the integral and determine the indefinite integral of the different terms using the power rule: ๏„ธ๐‘ฅโˆ’225๐‘ฅโˆ’15๐‘ฅ=๏„ธ(๐‘ฅ+15)๐‘ฅ=๏„ธ๐‘ฅ๐‘ฅ+๏„ธ15๐‘ฅ=๏„ธ๐‘ฅ๐‘ฅ+15๏„ธ1๐‘ฅ=๐‘ฅ1+1+15๐‘ฅ+=๐‘ฅ2+15๐‘ฅ+.๏Šจ๏Šง๏Šฐ๏Šง๏ŠจddddddCC

Note that we would obtain a constant of integration for each part from the process of integration, but we can combine these into one constant, C.

This result is valid for all ๐‘ฅโ‰ 15, since we require the integrand and integral to be continuous and well defined.

Now, letโ€™s consider an example where we determine the indefinite integral of a rational function with negative powers of ๐‘ฅ using the properties of linearity and the power rule for integrals.

Example 8: Finding the Integration of a Function Using the Power Rule for Integration with a Negative Exponent

Determine ๏„ธ๏€ผโˆ’8+89๐‘ฅ+75๐‘ฅ๏ˆ๐‘ฅ๏Šจ๏Šฌd.

Answer

In this example, we will determine the indefinite integral of the rational function โˆ’8+89๐‘ฅ+75๐‘ฅ๏Šจ๏Šฌ.

Letโ€™s first rewrite the integrand as powers of ๐‘ฅ: โˆ’8+89๐‘ฅ+75๐‘ฅ=โˆ’8+89๐‘ฅ+75๐‘ฅ.๏Šจ๏Šฌ๏Šฑ๏Šจ๏Šฑ๏Šฌ

In order to determine the integral, we will make use of the following properties of indefinite integrals: ๏„ธ(๐‘“(๐‘ฅ)+๐‘”(๐‘ฅ))๐‘ฅ=๏„ธ๐‘“(๐‘ฅ)๐‘ฅ+๏„ธ๐‘”(๐‘ฅ)๐‘ฅ,๏„ธ(๐‘Ž๐‘“(๐‘ฅ))๐‘ฅ=๐‘Ž๏„ธ๐‘“(๐‘ฅ)๐‘ฅ.ddddd

We will also make use of the power rule: ๏„ธ๐‘ฅ๐‘ฅ=๐‘ฅ๐‘›+1+,๐‘›โ‰ โˆ’1.๏Š๏Š๏Šฐ๏ŠงdC

Using the first property, we can split the given integral into three parts. We can then use the second property to take the appropriate factors outside the integral and determine the indefinite integral of the different terms using the power rule: ๏„ธ๏€ผโˆ’8+89๐‘ฅ+75๐‘ฅ๏ˆ๐‘ฅ=๏„ธ๏€ผโˆ’8+89๐‘ฅ+75๐‘ฅ๏ˆ๐‘ฅ=๏„ธโˆ’8๐‘ฅ+๏„ธ89๐‘ฅ๐‘ฅ+๏„ธ75๐‘ฅ๐‘ฅ=โˆ’8๏„ธ1+89๏„ธ๐‘ฅ๐‘ฅ+75๏„ธ๐‘ฅ๐‘ฅ=โˆ’8๐‘ฅ+89๏€พ๐‘ฅโˆ’2+1๏Š+75๏€พ๐‘ฅโˆ’6+1๏Š+=โˆ’8๐‘ฅ+89๏€พ๐‘ฅโˆ’1๏Š+75๏€พ๐‘ฅโˆ’5๏Š+=โˆ’8๐‘ฅโˆ’89๐‘ฅโˆ’725๐‘ฅ+๐ถ.๏Šจ๏Šฌ๏Šฑ๏Šจ๏Šฑ๏Šฌ๏Šฑ๏Šจ๏Šฑ๏Šฌ๏Šฑ๏Šจ๏Šฑ๏Šฌ๏Šฑ๏Šจ๏Šฐ๏Šง๏Šฑ๏Šฌ๏Šฐ๏Šง๏Šฑ๏Šง๏Šฑ๏Šซ๏ŠซdddddddCC

Note that we would obtain a constant of integration for each part from the process of integration, but we can combine these into one constant, C.

This result is valid for all ๐‘ฅโ‰ 0, since we require the integrand and integral to be continuous and well defined.

In the next example, we will determine the indefinite integral of a function with roots and negative exponents using the properties of linearity and the power rule for integrals.

Example 9: Finding the Integration of a Function Using the Power Rule for Integration with Roots and Negative Exponents

Determine ๏„ธ๏€ผโˆ’โˆš๐‘ฅ+8+9๐‘ฅ๏ˆ๐‘ฅ๏Šจd.

Answer

In this example, we will determine the indefinite integral of the function โˆ’โˆš๐‘ฅ+8+9๐‘ฅ๏Šจ.

Letโ€™s first rewrite the integrand as powers of ๐‘ฅ using โˆš๐‘ฅ=๐‘ฅ๏Ž ๏Žก: โˆ’โˆš๐‘ฅ+8+9๐‘ฅ=8โˆ’๐‘ฅ+9๐‘ฅ,๏Šจ๏Šฑ๏Šจ๏Ž ๏Žก for ๐‘ฅ>0. In order to determine the integral, we will make use of the following properties of indefinite integrals: ๏„ธ(๐‘“(๐‘ฅ)+๐‘”(๐‘ฅ))๐‘ฅ=๏„ธ๐‘“(๐‘ฅ)๐‘ฅ+๏„ธ๐‘”(๐‘ฅ)๐‘ฅ,๏„ธ(๐‘Ž๐‘“(๐‘ฅ))๐‘ฅ=๐‘Ž๏„ธ๐‘“(๐‘ฅ)๐‘ฅ.ddddd

We will also make use of the power rule: ๏„ธ๐‘ฅ๐‘ฅ=๐‘ฅ๐‘›+1+,๐‘›โ‰ โˆ’1.๏Š๏Š๏Šฐ๏ŠงdC

Using the first property, we can split the given integral into three parts. We can then use the second property to take the appropriate factors outside the integral and determine the indefinite integral of the different terms using the power rule: ๏„ธ๏€ผโˆ’โˆš๐‘ฅ+8+9๐‘ฅ๏ˆ๐‘ฅ=๏„ธ๏€ฝ8โˆ’๐‘ฅ+9๐‘ฅ๏‰๐‘ฅ=๏„ธ8๐‘ฅ+๏„ธโˆ’๐‘ฅ๐‘ฅ+๏„ธ9๐‘ฅ๐‘ฅ=8๏„ธ1๐‘ฅโˆ’๏„ธ๐‘ฅ๐‘ฅ+9๏„ธ๐‘ฅ๐‘ฅ=8๐‘ฅโˆ’๏‚๐‘ฅ+1๏Ž+9๏€พ๐‘ฅโˆ’2+1๏Š+=8๐‘ฅโˆ’๏‚๐‘ฅ๏Ž+9๏€พ๐‘ฅโˆ’1๏Š+=8๐‘ฅโˆ’23๐‘ฅโˆ’9๐‘ฅ+=8๐‘ฅโˆ’23โˆš๐‘ฅโˆ’9๐‘ฅ+.๏Šจ๏Šฑ๏Šจ๏Šฑ๏Šจ๏Šฑ๏Šจ๏Šฐ๏Šง๏Šง๏Šจ๏Šฑ๏Šจ๏Šฐ๏Šง๏Šฉ๏Šจ๏Šฑ๏Šง๏Šฑ๏Šง๏ŠฉddddddddCCCC๏Ž ๏Žก๏Ž ๏Žก๏Ž ๏Žก๏Ž ๏Žก๏Žข๏Žก๏Žข๏Žก

Note that we would obtain a constant of integration for each part from the process of integration, but we can combine these into one constant, C.

This result is valid for all ๐‘ฅ>0, since we require the integrand and integral to be continuous and well defined, and the square root is only defined for nonnegative numbers.

In the last example, we will determine the indefinite integral of a function involving fractional exponents by using factorization, the properties of linearity, and the power rule for integrals.

Example 10: Finding the Integration of a Function Using Factorization

Determine ๏„ธ36๐‘ฅโˆ’49โˆš๐‘ฅ(6๐‘ฅ+7)๐‘ฅ๏Šจ๏Žงd.

Answer

In this example, we will determine the indefinite integral of the function 36๐‘ฅโˆ’49โˆš๐‘ฅ(6๐‘ฅ+7)๏Šจ๏Žง.

We note that the numerator in the integrand can be written as the difference of two squares, as 36๐‘ฅโˆ’49=(6๐‘ฅ)โˆ’7=(6๐‘ฅ+7)(6๐‘ฅโˆ’7)๏Šจ๏Šจ๏Šจ. Using this, we can simplify the integrand and rewrite the remaining terms as powers of ๐‘ฅ using ๏Žง๏Ž ๏Žงโˆš๐‘ฅ=๐‘ฅ: 36๐‘ฅโˆ’49โˆš๐‘ฅ(6๐‘ฅ+7)=(6๐‘ฅ+7)(6๐‘ฅโˆ’7)โˆš๐‘ฅ(6๐‘ฅ+7)=6๐‘ฅโˆ’7โˆš๐‘ฅ=(6๐‘ฅโˆ’7)๐‘ฅ=6๐‘ฅโˆ’7๐‘ฅ,๏Šจ๏Žง๏Žง๏Žง๏Žช๏Ž ๏Žง๏Žฆ๏Žง๏Žช๏Ž ๏Žง for ๐‘ฅ>0. In order to determine the integral, we will make use of the following properties of indefinite integrals: ๏„ธ(๐‘“(๐‘ฅ)+๐‘”(๐‘ฅ))๐‘ฅ=๏„ธ๐‘“(๐‘ฅ)๐‘ฅ+๏„ธ๐‘”(๐‘ฅ)๐‘ฅ,๏„ธ(๐‘Ž๐‘“(๐‘ฅ))๐‘ฅ=๐‘Ž๏„ธ๐‘“(๐‘ฅ)๐‘ฅ.ddddd

We will also make use of the power rule: ๏„ธ๐‘ฅ๐‘ฅ=๐‘ฅ๐‘›+1+,๐‘›โ‰ โˆ’1.๏Š๏Š๏Šฐ๏ŠงdC

Using the first property, we can split the given integral into two parts. We can then use the second property to take the appropriate factors outside the integral and determine the indefinite integral of the different terms using the power rule: ๏„ธ36๐‘ฅโˆ’49โˆš๐‘ฅ(6๐‘ฅ+7)๐‘ฅ=๏„ธ๏€ฝ6๐‘ฅโˆ’7๐‘ฅ๏‰๐‘ฅ=๏„ธ6๐‘ฅ๐‘ฅ+๏„ธโˆ’7๐‘ฅ๐‘ฅ=6๏„ธ๐‘ฅ๐‘ฅโˆ’7๏„ธ๐‘ฅ๐‘ฅ=6๏‚๐‘ฅ+1๏Žโˆ’7๏‚๐‘ฅ+1๏Ž+=6๏‚๐‘ฅ๏Ž+7๏‚๐‘ฅ๏Ž+=6๏8๐‘ฅ15๏+7๏8๐‘ฅ7๏+=16๐‘ฅ5โˆ’8๐‘ฅ+.๏Šจ๏Šฐ๏Šง๏Šญ๏Šฎ๏Šฐ๏Šง๏Šฑ๏Šง๏Šฎ๏Šง๏Šซ๏Šฎ๏Šญ๏Šฎ๏Žง๏Žฆ๏Žง๏Žช๏Ž ๏Žง๏Žฆ๏Žง๏Žช๏Ž ๏Žง๏Žฆ๏Žง๏Žช๏Ž ๏Žง๏Žฆ๏Žง๏Žช๏Ž ๏Žง๏Ž ๏Žค๏Žง๏Žฆ๏Žง๏Ž ๏Žค๏Žง๏Žฆ๏Žง๏Ž ๏Žค๏Žง๏Žฆ๏ŽงddddddCCCC

Note that we would obtain a constant of integration for each part from the process of integration, but we can combine these into one constant, C.

This result is valid for all ๐‘ฅ>0, since we require the integrand and integral to be continuous and well defined, and the 8th root is only defined for nonnegative numbers.

Letโ€™s finish by considering the key points that we covered in this explainer.

Key Points

In order to determine the indefinite integrals of functions involving different powers of ๐‘ฅ including polynomial, reciprocal, and radical functions, we make use of the following:

  • The linearity property of integrals: ๏„ธ(๐‘Ž๐‘“(๐‘ฅ)+๐‘๐‘”(๐‘ฅ))๐‘ฅ=๐‘Ž๏„ธ๐‘“(๐‘ฅ)๐‘ฅ+๐‘๏„ธ๐‘”(๐‘ฅ)๐‘ฅ,ddd for ๐‘Ž,๐‘โˆˆโ„.
  • The power rule for integration: ๏„ธ๐‘ฅ๐‘ฅ=๐‘ฅ๐‘›+1+,๐‘›โ‰ โˆ’1.๏Š๏Š๏Šฐ๏ŠงdC

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