# Thread: Today's calculation of integral #1

1. ## Today's calculation of integral #1

Find $\int_0^1x\ln(1+x)\,dx.$

2. Originally Posted by Krizalid
Find $\int_0^1x\ln(1+x)\,dx.$
$\int_0^1 x\ln (x+1) dx = \int_0^1 \left[ x\cdot \sum_{n=0}^{\infty} \frac{(-1)^n x^n}{n} \right] dx= \sum_{n=0}^{\infty} \left[ \int_0^1 \frac{(-1)^n x^{n+1}}{n} \right]$
Now just simplify.

3. Originally Posted by Krizalid
Find $\int_0^1x\ln(1+x)\,dx.$
by parts u=ln(x+1) dv=xdx

$\frac{1}{2}x^2ln(x+1)|_{0}^{1}-\frac{1}{2}\int_{0}^{1}\frac{x^2}{x+1}dx$

rewriting the integrand we get

$\frac{ln(2)}{2}-\frac{1}{2}\int_{0}^{1}\frac{x^2-1+1}{x+1}dx$

$\frac{ln(2)}{2}-\frac{1}{2}\int_{0}^{1}\frac{(x-1)(x+1)+1}{x+1}dx=\frac{ln(2)}{2}-\frac{1}{2}\left[ \int_{0}^{1}(x-1)dx+\int_{0}^{1}\frac{1}{x+1}dx \right]$

$\frac{ln(2)}{2}-\frac{1}{2}[\frac{1}{2}x^2-x+ln(x+1)|_{0}^{1}]$

$\frac{ln(2)}{2}-\frac{1}{2}[\frac{1}{2}-1+ln(2)]=\frac{1}{4}$

4. I knew Hacker would use series, I was thinkin' the same too, even I like double integration tricks:

$\int_0^1 {x\ln (1 + x)\,dx} = \int_0^1 \!{\int_0^x {\frac{x}
{{1 + y}}\,dy} \,dx} .$

Now by reversing integration order the rest is routine. (And yes, the answer is $\frac14.$)