Apostol (6.22.10): Finding $\int \frac{\arcsin x}{x^2} dx$ I'm trying to solve another integral from Apostol (Chapter 6, Section 6.22, Question 10) which says to show the following:
$$
\int \frac {\arcsin x}{x^2}dx = \log|{\frac {1-\sqrt{1-x^2}}{x}}| - \frac {\arcsin x}{x} +C
$$
I have tried the following:
With $\int \frac {\arcsin x}{x^2}dx$ I started using integration by parts, choosing $u = \arcsin x $ and $dv = 1/x$ which lead to
$$
\int \frac {\arcsin x}{x^2}dx = -\frac{\arcsin x}{x} + \int \frac{dx}{x\sqrt{1-x^2}}
$$
$$
= -\frac{\arcsin x}{x} + \int \frac{x.dx}{x^2.sqrt{1-x^2}}
$$
With the Integral on the RHS, I set $u^2 = 1 - x^2$ which implied $x^2 = 1- u^2$ and $xdx = -u. du$. This meant
$$
\int \frac{xdx}{x^2\sqrt{1-x^2}} = \int \frac{-u du}{u (1-u^2)} = \int \frac{-du}{ 1-u^2} = -\int \frac{du}{1-u^2}
$$
For the integral $\int \frac{du}{1-u^2}$, I set $u = \sin \theta$ which meant $du = \cos\theta. d\theta$ which implied 
$$
\int \frac{du}{1-u^2} = \int \frac{\cos\theta}{1 -\sin^2\theta}d\theta = \int\sec\theta\,d\theta = \log|\sec \theta + \tan \theta| = \log|\frac{1+\sin \theta}{\cos \theta}|
$$
Now, since $u = \sin \theta$ in the last substitution and $\sin^2\theta + \cos^2\theta = 1$ it follows that $\cos\theta = \sqrt{1-u^2}$ which means
$$
\log|\frac{1+\sin \theta}{\cos \theta}| = \log |\frac{1+u}{\sqrt{1-u^2}}|
$$
$$
= \log|\frac{1+u}{\sqrt{1-u}\sqrt{1+u}}|
$$
$$
= \log|\frac{\sqrt{1+u}}{\sqrt{1-u}}| = \frac{1}{2}\log|\frac{1+u}{1-u}|
$$
That is
$$
\int \frac{du}{1-u^2} =\frac{1}{2}\log|\frac{1+u}{1-u}|
$$
Now, in the first substitution $u^2 = 1 - x^2$ implies $u = \sqrt{1-x^2}$ which means
$$
\frac{1}{2}\log|\frac{1+u}{1-u}| = \frac{1}{2}\log|\frac{1+ \sqrt{1-x^2}}{1 - \sqrt{1-x^2}}|
$$
$$
= \frac{1}{2}\log|\frac{(1+\sqrt{1-x^2})(1+\sqrt{1-x^2})}{(1 - \sqrt{1-x^2})(1+\sqrt{1-x^2})}|
$$
$$
= \frac{1}{2}\log|\frac{2 - x^2 +2\sqrt{1-x^2}}{x^2}|
$$
$$
= \log|\frac{1+\sqrt{1-x^2}}{x}|
$$
That is
$$
\int \frac{xdx}{x^2\sqrt{1-x^2}} = - \log|\frac{1+\sqrt{1-x^2}}{x}|
$$
with respect to the original integral means
$$
\int \frac {\arcsin x}{x^2}dx = -\frac{\arcsin x}{x} -  \log|\frac{1+\sqrt{1-x^2}}{x}| + C
$$
which is not the answer as stated in Apostol's book. I've probably made some error in the calculations which I just haven't been able to find, so if someone can point out the error in my calculations that would be appreciated
 A: Note that $(1+\sqrt{1-x^2})(1-\sqrt{1-x^2}) = 1 - (1-x^2) = x^2$, so
$$\log\left|\dfrac{1 + \sqrt{1-x^2}}{x}\right| = \log \left|\dfrac{x}{1-\sqrt{1-x^2}}\right| = - \log \left|\dfrac{1-\sqrt{1-x^2}}{x}\right|$$
You and Apostol have the same answer after all.
A: A much more computationally straightforward approach follows.  The integrand suggests the substitution $$\theta = \sin^{-1} x, \quad x = \sin \theta, \quad dx = \cos \theta \, d\theta,$$ giving $$I = \int \frac{\sin^{-1} x}{x^2} \, dx = \int \theta \csc \theta \cot \theta \, d\theta.$$  Then integration by parts with the choice $u = \theta$, $du = d\theta$, $dv = \csc\theta \cot\theta \, d\theta$, $v = -\csc \theta$, yields $$\begin{align*} I &= -\theta \csc \theta + \int \csc \theta \, d\theta \\ &= -\theta \csc \theta + \int \frac{\csc^2 \theta + \csc\theta \cot\theta}{\csc \theta + \cot \theta} \, d\theta.\end{align*}$$  Now with the choice $t = \csc\theta + \cot\theta$, $dt = -(\csc^2 \theta + \csc\theta \cot\theta) \, d\theta$, we obtain $$\begin{align*}I &= -\theta \csc\theta - \log|\csc\theta + \cot\theta| + C \\ &= -\frac{\sin^{-1} x}{x} - \log \left| \frac{1 + \sqrt{1-x^2}}{x} \right| + C.\end{align*}$$  Robert Israel's answer then shows the equivalence of this antiderivative to the claimed result.
