# Math Help - integration by parts - using a previous integration

1. ## integration by parts - using a previous integration

Hello

I am working on a problem which involves integrating $\int \frac{x^3}{1+x} dx$ [with limits 0, 1 (don't know how to do limits in latex)].

So I managed to integrate using substitution as follows:

let u = 1+x, du/dx = 1, limits: u = 1+1 = 2 and u = 1+0=1

x = u-1.

So get:
$\int \frac{(u - 1)^3}{u} du$ [with different limits 1, 2]

I expand out:
$\int u^2 - 3u + 3 - \frac{1}{u} du$

and then:
$\frac{u^3}{3} - \frac{3u^2}{2} + 3u - ln u$

and I get a result of 5/6 - ln2.

I am then asked to integrate this: $\int x^2 ln(1+x)$ with limits 0 to 1.

So I start integrating (by parts) and notice that second integration bit - I have this:

= $ln(1+x) . \frac{x^3}{3} - \frac{1}{3} \int \frac{x^3}{1+x}$

That last bit, I already integrated above.

So my question is, how do I combine the two. My start part of integration by parts and then the minus - and here use my previous result.

The previous result had different limits - so I presume I cant just use the result. Do I just substitute back in 1+x for the u above like this:

$ln(1+x) . \frac{x^3}{3} - \frac{1}{3}(\frac{(1+x)^3}{3} - \frac{3(1+x)^2}{2} + 3(1+x) - ln (1+x))$ - then use my limits 0, 1 as asked for this integration?

Angus

2. ## Re: integration by parts - using a previous integration

You can say that $\displaystyle \frac{x^3}{1+x} = x^2-x+1-\frac{1}{1+x}$

3. ## Re: integration by parts - using a previous integration

What you got is this: $\int_0^1 x^2 \ln (1+x) \, dx = \frac{x^3}{3} \cdot \ln (1+x) \Big\vert_0^1 - \frac{1}{3} \int_0^1 \frac{x^3}{1+x} \, dx$?

You already know the value of $\int_0^1 \frac{x^3}{1+x} \, dx$, just divide it by 3 and subtract, yielding:

$\int_0^1 x^2 \ln (1+x) \, dx = \frac{1^3}{3} \cdot \ln 2 - \frac{1}{3} \cdot \left( \frac{5}{6} - \ln 2 \right) = \frac{2 \ln 2}{3} - \frac{5}{18}$