What is the purpose of rearranging this integration by parts expression?

Question

In my textbook they give this series of steps:

$${d\over dx}(u(x)v(x))=u'(x)v(x)+u(x)v'(x)$$

Then integrating both sides:

$$u(x)v(x)=\int(u'(x)v(x)+u(x)v'(x))dx$$

Then the textbook rearranges the expression, and it only uses the rearranged version throughout the rest of the book (well, at least throughout the rest of the section).

$$\int u(x)v'(x)dx=u(x)v(x)-\int v(x)u'(x)dx$$

It doesn't explain in the book why this is the preferred version. So my question is: Why is it better?

I looked around online and I couldn't find an explanation, perhaps it isn't anywhere because it's a ridiculous question.

Answer 1

It is convenient to use the rearranged form in order to integrate some functions. For example, suppose we are trying to find the antiderivative of $xe^x$. This means we are trying to find $\int xe^x dx$. So, if we take $u=x$ and $v'(x)=e^x$, we get $u'(x)=1$ and $v(x)=e^x$ (ignoring the $+C$ for now), and so:

$\int xe^xdx=xe^x-\int 1\cdot e^xdx=xe^x-\int e^xdx=xe^x-e^x+C$.

Answer 2

It is often written as $\int u dv=uv-\int v du.$ For example $\int x^2 \cos x dx=\int x^2 d\sin x=x^2 \sin x-\int \sin x d(x^2)= x^2 \sin x-\int (\sin x)(2x dx) =$ $x^2 \sin x-\int 2x d(-\cos x)=$ $x^2 \sin x-(2x.(-\cos x)-\int (-\cos x)) d(2x)=$ $x^2 \sin x+2x.\cos x-2\sin x.$