Hello!

I have tried to solve these two integrals, but I get the wrong answer.

The first one is 4/x^2 - 4 (upper limit is 1, lower limit 0).

The correct answer is -ln2, but I get 2arctan(x/2) + c.

$\displaystyle \frac{4}{x^2-4}=\frac{1}{x-2}-\frac{1}{x+2}$ and thus this a a logarithmic integral: $\displaystyle \ln\left|\frac{x-2}{x+2}\right|_0^1=\ln\frac{1}{3}-\ln 1=-\ln 3$ , so imo both answers are incorrect.
The next one is ln^2 x/2x (upper limit e^2, lower limit 1).

I am not sure what the correct answer is, but either

a) 4/3 b) 1/4 c) 3/5 d) 7/6 e) 4/9

$\displaystyle \int\limits_1^{e^2}\frac{\ln^2x}{2x}=\frac{1}{6}\l eft[\ln^3x\right]_1^{e^2}=\frac{1}{6}(8-0)=\frac{4}{3}$

Tonio
I would appreciate your help,

best wishes

Amine