<s>[INST] What is the value of $ \int_{x}^{1} \arcsin \left( \frac{2t}{t^2+1} \right) \text{d}t $? Is this result true? Wolfram doesn't seem to be able to evaluate the definite integral in the allowed time. $$ \int_{x}^{1} \arcsin \left( \dfrac{2t}{t^2+1} \right) \text{d}t = \dfrac{\pi}{2} - 2x\arctan x - \log(2) + \log(1+x^2) $$
 [/INST] $$
\frac{2t}{1+t^2} = \frac{2\tan\theta}{1+\tan^2\theta} = \frac{2\tan\theta}{\sec^2\theta} = 2\sin\theta\cos\theta = \sin (2\theta).
$$
So
$$
\arcsin\left(\frac{2t}{1+t^2}\right) = 2\theta.
$$
$$
dt = \sec^2\theta\,d\theta
$$
As $t$ goes from $x$ to $1$, then $\theta=\arctan t$ goes from $\arctan x$ to $\pi/4$.
$$
\int \theta\Big(\sec^2\theta\,d\theta\Big) = \int \theta\,dv = \theta v- \int v\,d\theta
$$
and so $v=\tan\theta$, etc.
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