# Use integration by parts to express $\int f(x)dx$ in terms of $\int g(x)dx$

This is the first question from Single Variable Calculus 6th Edition Chapter 7, Section 2. (Authors: Hughes-Hallet, Gleason, McCallum, et al.)

I've formatted the title of this question to be in a more general form, but the question from the book is as follows:

Use integration by parts to express $$\int x^2e^xdx$$ in terms of: $$a. \int x^3e^xdx \qquad \qquad b. \int xe^xdx$$

The book also provides the solution in the back, but not the arithmetic to justify it.

1. (a) $$\frac {x^3e^x}{3} - \frac {1}{3} \int x^3e^xdx$$

(b) $$x^2e^x-2 \int xe^xdx$$

What I understand:

Integration by parts.

What I do not understand:

How they got that solution.

• Hint: apply parts with $u=x^2, dv=e^xdx$. Then apply parts with $u=e^x, dv=x^2dx$. – lulu Sep 21 '19 at 22:50
• Just to clarify, the book is basically asking that the anti-derivative of g(x) be constructed using u and dv from f(x)? And what comes before the anti-deriviative of g(x) as a result of integration by parts using f(x) doesn't matter? Thank you for the hint by the way, it helped a lot. – Evilscaught Sep 21 '19 at 23:28
• Well, I'm not sure of your notation. The version of parts I am using is $\int u\,dv=uv-\int v \,du$. – lulu Sep 21 '19 at 23:32
• The integration by parts notation I'm using is from the book, it is as follows: $\int uv'dx = uv - \int u'vdx$ But, I have previously seen the notation that you're using. – Evilscaught Sep 21 '19 at 23:37
• Well, that's more or less the same...as $dv=v'\,dx$ and $du=u'\,dx$. – lulu Sep 21 '19 at 23:39

Integration by parts just exploits the product rule for derivatives and the fundamental theorem of calculus: $$fg=\int (fg)'=\int f'g+\int fg'\implies \int f'g=fg-\int fg'$$.
So, $$(\dfrac{x^3e^x}3)'=x^2e^x+\dfrac {x^3e^x}3\implies \int x^2e^x=\dfrac {x^3e^x}3-\int\dfrac {x^3e^x}3$$.
$$b)$$ is done the same way.