Video: Finding the Integration of a Function Involving Logarithmic Functions by Using Integration by Substitution

Determine ∫ βˆ’(4/(9π‘₯(ln 9π‘₯)⁷) dπ‘₯) by using the substitution method.

03:58

Video Transcript

Determine the indefinite integral of negative four over nine π‘₯ multiplied by the natural logarithm of nine π‘₯ to the seventh power with respect to π‘₯ by using the substitution method.

Now, this looks like a very complicated integral to evaluate. We have a reciprocal, and in the denominator we have the product of an algebraic function and a natural logarithm function, which is raised to the seventh power. But fortunately, we’re given a clue. We’re told that we should answer this question using the substitution method. In the substitution method, we’re looking to change the variable, so that the integral becomes more straightforward to perform. What we’re looking for when we use the substitution method is for the integrand to be in the form 𝑔 prime of π‘₯ multiplied by 𝑓 of 𝑔 of π‘₯. That’s a function of a function multiplied by the derivative of the other function or at least some multiple of this.

Now, we noticed that the derivative with respect to π‘₯ of the natural logarithm of nine π‘₯ is one over π‘₯. And if we rewrite our integral slightly as negative four-ninths times the integral of one over π‘₯ multiplied by one over the natural logarithm of nine π‘₯ to the seventh power with respect to π‘₯, then we see that we do indeed have some function of the natural logarithm of nine π‘₯ multiplied by the derivative of the natural logarithm of nine π‘₯. Let’s see what happens then if we make this substitution will introduce a new variable 𝑒 to equal the natural logarithm of nine π‘₯.

We’ve already said that the derivative with respect to π‘₯ of the natural logarithm of nine π‘₯ is one over π‘₯. So we have d𝑒 by dπ‘₯ equals one over π‘₯. Remember that d𝑒 by dπ‘₯ is not a fraction. But it is equivalent to say that d𝑒 is equal to one over π‘₯ dπ‘₯. So looking at the second form that we wrote our integral in, we can replace one over π‘₯ dπ‘₯ with d𝑒 and we can replace the natural logarithm of nine π‘₯ with 𝑒. This gives us negative four-ninths of the integral of one over 𝑒 to the seventh power with respect to 𝑒 and we’ve, therefore, changed our integral fully from an integral in terms of π‘₯ to an integral in terms of 𝑒.

It’s also an integral that we can perform without any great difficulty. If we will call that one over 𝑒 to the seventh power can be expressed as 𝑒 to the power of negative seven and integrate powers of 𝑒 not equal to negative one, we increase the power by one and then divide by the new power. So we have negative four-ninths multiplied by one over negative six 𝑒 to the power of negative six. And we also need to include a constant of integration 𝑐, as this is an indefinite integral. We can perform some simplification at our answer that negative in the numerator will cancel with the negative in the denominator. And the four in the numerator can be divided by two to give two. And the six in the denominator can be divided by two to give three.

We, therefore, have to over 27𝑒 to the six power plus our constant of integration 𝑐. But we aren’t finished yet. This answer is in terms of 𝑒 and the original integral was in terms of π‘₯. As this is an indefinite integral, we must make sure we reverse our substitution. So we need to swap 𝑒 for its definition in terms of π‘₯. That’s the natural logarithms of nine π‘₯. And so we have a final answer. By using the substitution method, we found that the indefinite integral of negative four over nine π‘₯ multiplied by the natural logarithm of nine π‘₯ to the seventh power with respect to π‘₯ is equal to two over 27 multiplied by the natural logarithm of nine π‘₯ to the sixth power plus a constant of integration 𝑐.

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