<s>[INST] Solving $\int_{0}^{1} \dfrac{\ln(x+1)}{x^2+1} dx$ The integral is:
$$\int_{0}^{1} \dfrac{\ln(x+1)}{x^2+1} dx$$
This is a Putnam question from my calculus book. It was in the section about trig substitution. However, I see no way of how mixing a trig function with a natural logarithm would make it any easier. I've tried the substitution $x=\tan \theta$ and got $$\dfrac 12\int \ln(1+\tan \theta)^2 d \theta$$
However, I wasn't able to simplify it to any integrable form.
 [/INST] A substitution other than a trigonometric substitution can be used to evaluate this integral.
Using a so-called self-similar substitution of $u = \dfrac{1 - x}{1 + x}$, one has $x = \dfrac{1 - u}{1 + u}$. Thus $dx = -\dfrac{2}{(1 + u)^2} \, du$ and for the limits of integration when $x = 0, u = 1$ and when $x = 1, u = 0$. 
Noting that $1 + x = \dfrac{2}{1 + u}$ and $1 + x^2 = \dfrac{2(1 + u^2)}{(1 + u)^2}$ the integral becomes
\begin{align*}
I &= \int^1_0 \frac{\ln (1 + x)}{1+ x^2} \, dx\\
&= \int^1_0 \ln \left (\frac{2}{1 + u} \right ) \cdot \frac{(1 + u)^2}{2 (1 + u^2)} \cdot \frac{2}{(1 + u)^2} \, du\\
&= \int^1_0 \ln \left (\frac{2}{1 + u} \right ) \frac{du}{1 + u^2}\\
&= \int^1_0 \left [\ln 2 - \ln (1 + u) \right ] \frac{du}{1 + u^2}\\
&= \ln 2 \int^1_0 \frac{du}{1 + u^2} - \int^1_0 \frac{\ln (1 + u)}{1 + u^2} \, du\\
\Rightarrow 2I &= \ln 2 \cdot \Big{[}\tan^{-1} u \Big{]}^1_0
\end{align*}
giving
$$\int^1_0 \frac{\ln (1 + x)}{1 + x^2} \, dx = \frac{\pi}{8} \ln 2.$$
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