# Breaking a looping integral $\int\frac1x\ln(\frac1{x^2+1})dx$

Said integral being $$\int\frac1x\ln(\frac1{x^2+1})dx$$ I've tried solving it with normal integration by parts exploiting that $$\ln(\frac1{x^2+1})=(-1)\ln(x^2+1)$$ but no dice. I'm trying to find a way to shoehorn in $arctanx$ as a primitive of $\frac1{x^2+1}$ without making the rest unreasonably complicated.

• Do you expect to find a closed-form, elementary anti-derivative? – StackTD Jul 12 '18 at 11:49
• I'm actually just supposed to find the definite integral for 1<x<2. I tried to pose y=(starting integral) and find it indirectly but it just loops on itself with no usable coefficents. – Moss Jul 12 '18 at 11:51
• It's always useful to add this information and/or relevant context; sometimes you can find a definite integral without explicitly finding an anti-derivative. – StackTD Jul 12 '18 at 11:53

HINT: $$\ln\left(\frac1{x^2+1}\right) = -\ln \left(x^2+1\right)$$ $$-\int\frac{\ln \left(x^2+1\right)}{x}dx$$ Set $$u = -x^2 \implies dx = -\frac{1}{2x}\,du$$ $$-\frac{1}{2}\int\frac{\ln\left(1-u\right)}{u}du$$