Question

Integrate x/(1+xy) dy?

Answer 1

@SithsAndGiggles

Answer 2

Is \(x\) fixed, or are you computing a double integral?

Answer 3

It's a double integral x/(1+xy) dy dx from 0 to 1 and from 0 to 1.

Answer 4

\[I=\int_0^1\int_0^1\frac{x}{\color{green}{1+xy}}\,\color{red}{dy}\,dx\]
Integrating with respect to \(y\) first, you can fix \(x\) so that you treat it like a constant. Substitute \(\color{green}{u=1+xy}\), then take the differential (remember, \(x\) is fixed), which would be \(\color{blue}{du}=\color{blue}{x}\color{red}{\,dy}\).
\[I=\int_0^1\int_1^{1+x}\frac{x}{\color{green}u}\color{blue}{\frac{du}{x}}\,dx\]

Answer 5

But how to integrate ln abs(1+x) dx?

Answer 6

With respect to \(x\), you're integrating over the interval \([0,1]\). For \(x>-1\), you have \(\ln|1+x|=\ln(1+x)\), so you have
\[I=\int_0^1\ln(1+x)\,dx\]

Answer 7

Okay, thanks!