Integral of $\,\arcsin(x)/(x^2)$

Question

I did the integration by parts and got this expression, but then I am stuck on how to take it further. I tried substituting u=1-x^2, but then I had to do a partial fraction decomposition (which I did not take). Any hints or help would be very appreciated!

Try the substitution $u = \sqrt{1-x^2}$ in the last integral. — Robert Israel, Jan 24 '18 at 03:54

score 2 · Answer 1 · edited Sep 12 '20 at 18:24

2

Take $x = \sin u$ initially to get simple integral.

$$\int u \csc( u )\cot( u) du= -u\csc( u) + \int \csc (u) du\\ = -u \csc (u) +\ln|\cot( u )-\csc (u)| + c.$$

edited Sep 12 '20 at 18:24

amWhy

209,954

answered Jan 24 '18 at 04:09

King Tut

3,074

score 1 · Answer 2 · answered Jan 24 '18 at 04:08

Let $\sqrt{1-x^2}=t$.

Thus, $-\frac{2x}{2\sqrt{1-x^2}}dx=dt$ and $$\int\frac{1}{x\sqrt{1-x^2}}dx=-\int\left(\frac{1}{x\sqrt{1-x^2}}\cdot\frac{\sqrt{1-x^2}}{x}\right)dt=$$ $$=-\int\frac{1}{x^2}dt=-\int\frac{1}{1-t^2}dt=\frac{1}{2}\int\left(\frac{1}{t-1}-\frac{1}{t+1}\right)dt=$$ $$=\frac{1}{2}\ln\left|\frac{t-1}{t+1}\right|+C=\frac{1}{2}\ln\left|\frac{\sqrt{1-x^2}-1}{\sqrt{1-x^2}+1}\right|+C.$$

score -1 · Answer 3 · answered Aug 07 '19 at 11:15

-1

I wrote my answer for this question, click here to see the image.

answered Aug 07 '19 at 11:15

Ryan Su

1

Welcome to MSE. It is in your best interest that you type your questions (using MathJax) instead of posting links to pictures. – José Carlos Santos Aug 07 '19 at 11:36

Integral of $\,\arcsin(x)/(x^2)$

3 Answers3