# Question Video: Finding the Integration of a Function Involving Using Integration by Substitution and Integration by Parts Mathematics • Higher Education

Determine β« (2π^(π₯) π₯)/(3(π₯ + 1)Β²) dπ₯.

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### Video Transcript

Determine the indefinite integral of two π to the power of π₯ times π₯ over three times π₯ plus one all squared, evaluated with respect to π₯.

It might not be instantly obvious how we are going to evaluate this. However, thereβs no clear substitution we can make. And itβs certainly not something we can evaluate in our heads. So that tells us we might need to use integration by parts. Remember, this says the integral of π’ times dπ£ by dπ₯ is equal to π’π£ minus the integral of π£ times dπ’ by dπ₯. Now, if we compare this formula to our integrand, we see that we need to decide which function is π’. And which function is dπ£ by dπ₯. So how do we decide this? Well, our aim is to make sure that the second integral we get over here is a little simpler. We therefore want π’ to be a function that either becomes simpler when differentiated or helps us to simplify the integrand when multiplied by this π£. Now it isnβt hugely obvious what we should choose π’ to be. So a little trial and error might be in order.

Letβs rewrite our integrand as two-thirds π₯ times π to the power of π₯ times one over π₯ plus one squared. And in fact, letβs take the constant factor of two-thirds outside of the integral. Letβs try π’ equals π₯ times π to the power of π₯ and dπ£ by dπ₯ as being one over π₯ plus one all squared, which Iβve written as π₯ plus one to the power of negative two. To find dπ’ by dπ₯, weβre going to need to use the product rule. If we do, we see that dπ’ by dπ₯ is equal to π₯ times the derivative of π to the power of π₯ plus π to the power of π₯ times the derivative of π₯. Well, the derivative of π to the power of π₯ is π to the power of π₯. And the derivative of π₯ is one. So weβve obtained dπ’ by dπ₯ to be equal to π₯π to the power of π₯ plus π to the power of π₯. We can use the reverse of the chain rule to work out the antiderivative of π₯ plus one to the power of negative two. Itβs negative π₯ plus one to the negative one. Rewriting π₯ plus one to the power of negative one as one over π₯ plus one. And we can substitute everything we have into our formula for integration by parts.

Now over here, this next step is important. We factor by π to the power of π₯. And then we can see we can divide through by π₯ plus one. And our second integrand becomes negative π to the π₯. We take out that factor of negative one. And we know that the integral of π of the power of π₯ is simply π to the power of π₯. And so we have two-thirds of π₯ times π to the power of π₯ over negative π₯ plus one plus π to the power of π₯ plus π. We rewrite this first fraction slightly and then multiply the numerator and the denominator of π to the power of π₯ by π₯ plus one. So that we can add these fractions. And we see that the sum of these fractions is negative π₯ times π to the power of π₯ plus π₯ times π to the power of π₯, which is zero, plus π to the power of π₯ over π₯ plus one.

Our final step is to distribute the parentheses. And we get two π to the power of π₯ over three times π₯ plus one. And Iβve changed our constant of integration to capital πΆ. Since our original constant of integration was multiplied by two-thirds.