Question Video: Integration of a Power Function and a Reciprocal Function | Nagwa Question Video: Integration of a Power Function and a Reciprocal Function | Nagwa

# Question Video: Integration of a Power Function and a Reciprocal Function Mathematics • Third Year of Secondary School

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Determine β« (8π₯βΉ + 4/π₯) dπ₯.

02:25

### Video Transcript

Determine the integral of eight multiplied by π₯ to the ninth power plus four divided by π₯ with respect to π₯.

We start by recalling that the integral of the sum of two functions π and π with respect to π₯ is equal to the integral of π with respect to π₯ plus the integral of π with respect to π₯. So, we can use this to split our integral into two separate integrals. We get the integral of eight multiplied by π₯ to the ninth power with respect to π₯ plus the integral of four divided by π₯ with respect to π₯.

Next, we recall that if π is not equal to negative one, then the integral of some constant π multiplied by π₯ to the πth power with respect to π₯ is equal to π divided by π plus one multiplied by π₯ to the power of π plus one plus our constant of integration, π.

We can use this to integrate eight multiplied by π₯ to the ninth power with respect to π₯. Weβll set our exponent, π, equal to nine and our coefficient, π, equal to eight. Doing this gives us eight divided by nine plus one multiplied by π₯ to the power of nine plus one plus a constant of integration we will call π one.

Now, we recall that the integral of some constant π divided by π₯ with respect to π₯ is equal to π multiplied by the natural logarithm of the absolute value of π₯ plus a constant of integration, π. We can use this to integrate four divided by π₯ with respect to π₯. Weβll set our coefficient, π, to be equal to four. This gives us four multiplied by the natural logarithm of the absolute value of π₯ plus a constant of integration we will call π two.

Weβre now ready to start simplifying. We have nine plus one is equal to 10. So, our first term is eight divided by 10 multiplied by π₯ to the 10th power. We have the both π one and π two, our constants of integration, so we can combine both of these into a new constant, which we will call π. This gives us eight over 10 multiplied by π₯ to the 10th power plus π plus four multiplied by the natural logarithm of the absolute value of π₯. Finally, we can simplify eight divided by 10 to just be four divided by five.

Therefore, we have shown that the integral of eight multiplied by π₯ to the ninth power plus four over π₯ with respect to π₯ is equal to four multiplied by π₯ to the 10th power divided by five plus four multiplied by the natural logarithm of the absolute value of π₯ plus a constant of integration, π.

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