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. $$ \int_{x}^{1} \arcsin \left( \dfrac{2t}{t^2+1} \right) \text{d}t = \dfrac{\pi}{2} - 2x\arctan x - \log(2) + \log(1+x^2) $$
 A: To find out whether it is true, differentiate, using the Fundamental Theorem of Calculus. Then use the fact that the integral on the left is $0$ at $x=1$.
A: HINTS:
Make the substitution $t=\tan\theta$, and hence $\mathrm{d}t = \sec^2\theta~\mathrm{d}\theta$.
The identities $1+\tan^2\theta \equiv \sec^2\theta$ and $2\sin\theta\cos\theta \equiv \sin \theta$ will also be required.
You will be able to reduce the indefinite integral to
$$\int 2\theta\sec^2\theta~\mathrm{d}\theta$$
You can integrate this by parts by putting $u=2\theta$ and $\mathrm{d}v = \sec^2\theta~\mathrm{d}\theta$.
Once you have integrated $2\theta\sec^2\theta$, you will need to make the reverse substitution $\theta=\arctan t$ and then apply the identity
$$\cos(\arctan t) \equiv \frac{1}{\sqrt{1+t^2}}$$
Finally, you will need to apply laws of logs to $\ln((1+t^2)^{-1/2})$.
NOTE: 
Be careful with the domains of definition of the inverse trigonometric functions.
A: $$
\frac{2t}{1+t^2} = \frac{2\tan\theta}{1+\tan^2\theta} = \frac{2\tan\theta}{\sec^2\theta} = 2\sin\theta\cos\theta = \sin (2\theta).
$$
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.
