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I know I have to make a u substitution and then do integration by parts. $$\int x\ln(1+x)dx$$

$ u = 1 + x$

$du = dx$

$$\int (u-1)(\ln u)du$$

$$\int u \ln u du - \int \ln u du$$

I will solve the $\ln u$ problem first since it will be easier

$$ \int \ln u du$$

$u = \ln u$

$du = 1/u$

$dz = du$

$z = u$

$$-(u\ln u - u)$$

Now I will do the other part.

$$\int u \ln u du$$

$u = \ln u$ $du = 1/u$

$dz = udu$ $z = u^2 / 2$

$$\frac {u^2 \ln u}{2} - \int u/2$$

$$\frac {u^2 \ln u}{2} - \frac{1}{2} \int u$$

$$\frac {u^2 \ln u}{2} - \frac{u^2}{2} $$

Now add the other part.

$$\frac {u^2 \ln u}{2} - \frac{u^2}{2} -u\ln u + u $$

Now put u back in terms of x.

$$\frac {(1+x)^2 \ln (1+x)}{2} - \frac{(1+x)^2}{2} -(1+x)\ln (1+x) + (1+x) $$

This is wrong and I am not sure why.

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    $\begingroup$ I'd like to commend you on the work you included. I know it takes time, but it is really helpful for us to track down the problem. $\endgroup$
    – rschwieb
    Jun 1 '12 at 13:27
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Comment for Jordan: You should not read the solution below, it is somewhat non-standard and at this stage you need to think in terms of standard approaches.

We use integration by parts, $u=\ln(1+x)$, $dv=x\,dx$. So $du =\frac{dx}{1+x}$.
Now we do something cute with $v$. Any antiderivative of $x$ will do. Instead of boring old $\frac{x^2}{2}$, we can take $$v=\frac{x^2}{2}-\frac{1}{2}=\frac{1}{2}(x+1)(x-1).$$ Thus $$\int x\ln(1+x)\,dx=\left(\frac{x^2}{2}-\frac{1}{2}\right)\ln(1+x)-\int \frac{1}{2}(x-1)\,dx.$$

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  • $\begingroup$ So "cute" :) +1 $\endgroup$
    – user399078
    Jan 29 '17 at 15:29
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Try taking the derivative of your solution to "the other part" ($\frac {u^2 \ln u}{2} - \frac{u^2}{2} $) and compare to where you started ($u\ln u$). This approach should serve as a general rule for checking integrals (that you don't have memorized, at least).

As another general rule, try not to reuse variables, e.g., in your line $u = \ln u\space \text{d}u = 1/u$; it can lead to mistakes as you forget which $u$ is which. $u \ne 1/u$, which is a clue (and no, $\ln u\space \text{d}u \ne 1/u$.)

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First of all I want to caution you about overusing $u$ as two different variables. When I tried it, I used $t$ to stubstitute at the outset instead, just to avoid any accidental confusion. (This confusion could affect either you, or whoever is reading your work.)

Secondly there is a mistake here: $\frac {u^2 \ln u}{2} - \frac{u^2}{2}$. this should be $\frac {u^2 \ln u}{2} - \frac{u^2}{4}$, because before you integrated $u$, there was already a 1/2 outside.

So I agree with $\int x\ln(1+x) dx=\frac {(1+x)^2 \ln (1+x)}{2} - \frac{(1+x)^2}{4} -(1+x)\ln (1+x) + (1+x)+C$.

You can also omit the 1 if you like, by combining it into $C$.

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