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This is from a list of problems from a local university's Integration Bee. I have no idea how to do it, but I thought maybe someone here can explain it to me.

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

The answer they have is $\frac{\pi \ln(2)}{8}$, but no clue how they got there.

Thanks!

BlahK
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1 Answers1

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Let $$I = \int_{0}^{1}\frac{\ln(1+x)}{1+x^2}dx\;,$$ Put $$x=\frac{1-t}{1+t} = -\left(\frac{t-1}{t+1}\right)=\frac{2}{t+1}-1$$

So we get $$dx=-\frac{2}{(1+t)^2}dt$$ and changing Limits ,We get

$$I = -\int_{1}^{0}\frac{\ln\left(\frac{2}{1+t}\right)}{2(1+t^2)}\cdot \frac{2}{(1+t)^2}\cdot (1+t)^2dt = \int_{0}^{1}\frac{\ln 2-\ln(1+t)}{1+t^2}dt$$

So we get $$I = \ln 2\int_{0}^{1}\frac{1}{1+t^2}dt-I \Rightarrow 2I = \frac{\pi}{4}\cdot \ln 2$$

So we get $$I = \int_{0}^{1}\frac{\ln(1+x)}{1+x^2}dx = \frac{\pi}{8}\cdot \ln 2$$

juantheron
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