Video: Finding the Integration of a Function Involving Using Integration by Substitution and Integration by Parts

Determine ∫ (2𝑒^(π‘₯) π‘₯)/(3(π‘₯ + 1)Β²) dπ‘₯.

03:09

Video Transcript

Determine the indefinite integral of two 𝑒 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, there’s no clear substitution we can make. And it’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 𝑒 times d𝑣 by dπ‘₯ is equal to 𝑒𝑣 minus the integral of 𝑣 times d𝑒 by dπ‘₯. Now, if we compare this formula to our integrand, we see that we need to decide which function is 𝑒. And which function is d𝑣 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 𝑒 to be a function that either becomes simpler when differentiated or helps us to simplify the integrand when multiplied by this 𝑣. Now it isn’t hugely obvious what we should choose 𝑒 to be. So a little trial and error might be in order.

Let’s rewrite our integrand as two-thirds π‘₯ times 𝑒 to the power of π‘₯ times one over π‘₯ plus one squared. And in fact, let’s take the constant factor of two-thirds outside of the integral. Let’s try 𝑒 equals π‘₯ times 𝑒 to the power of π‘₯ and d𝑣 by dπ‘₯ as being one over π‘₯ plus one all squared, which I’ve written as π‘₯ plus one to the power of negative two. To find d𝑒 by dπ‘₯, we’re going to need to use the product rule. If we do, we see that d𝑒 by dπ‘₯ is equal to π‘₯ times the derivative of 𝑒 to the power of π‘₯ plus 𝑒 to the power of π‘₯ times the derivative of π‘₯. Well, the derivative of 𝑒 to the power of π‘₯ is 𝑒 to the power of π‘₯. And the derivative of π‘₯ is one. So we’ve obtained d𝑒 by dπ‘₯ to be equal to π‘₯𝑒 to the power of π‘₯ plus 𝑒 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. It’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 𝑒 to the power of π‘₯. And then we can see we can divide through by π‘₯ plus one. And our second integrand becomes negative 𝑒 to the π‘₯. We take out that factor of negative one. And we know that the integral of 𝑒 of the power of π‘₯ is simply 𝑒 to the power of π‘₯. And so we have two-thirds of π‘₯ times 𝑒 to the power of π‘₯ over negative π‘₯ plus one plus 𝑒 to the power of π‘₯ plus 𝑐. We rewrite this first fraction slightly and then multiply the numerator and the denominator of 𝑒 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 𝑒 to the power of π‘₯ plus π‘₯ times 𝑒 to the power of π‘₯, which is zero, plus 𝑒 to the power of π‘₯ over π‘₯ plus one.

Our final step is to distribute the parentheses. And we get two 𝑒 to the power of π‘₯ over three times π‘₯ plus one. And I’ve changed our constant of integration to capital 𝐢. Since our original constant of integration was multiplied by two-thirds.

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