Find indefinite integral $\int \frac{\arcsin (x)}{x^2}\, \mathrm{d}x$

Question

$$\int \frac{\arcsin (x)}{x^2}\, \mathrm{d}x$$ $$\frac {1}{x^2\sqrt{1-x^2}}+2\int \frac{1}{x^3\sqrt{1-x^2}}\, \mathrm{d}x$$

I try to integrate by parts method, but its doesnt want to be solved. I try to substitue $x=\sin u \mathrm{d}x=\cos u$ but failed somewhere

Answer 1

Let $u=\arcsin x$ and $dv=\frac{dx}{x^2}$. Then we will get (see Michael Burr answer. He used partial fractions and mine is trigo substitution) $$\int \frac{\arcsin x}{x^2}dx=-\frac{\arcsin x}{x}+\int\frac{dx}{x\sqrt{1-x^2}}.$$

Let $x=\sin \theta$. Then $dx=\cos\theta\ d\theta$ and $$\sqrt{1-x^2}=\cos\theta.$$ Thus, \begin{align} \int \frac{\arcsin x}{x^2}dx&=-\frac{\arcsin x}{x}+\int\frac{\cos\theta\ d\theta}{\sin\theta\cos\theta}\\ &=-\frac{\arcsin x}{x}+\int \csc\theta\ d\theta\\ &=-\frac{\arcsin x}{x}-\ln|\csc\theta+\cot\theta|+C\\ &=-\frac{\arcsin x}{x}-\ln\left| \frac{1+\sqrt{1-x^2}}{x}\right|+C\\ &=-\frac{\arcsin x}{x}-\ln\left|1+\sqrt{1-x^2}\right|+\ln|x|+C \end{align}

Answer 2

Let $u = \arcsin x$ and life will be good. (Change $\cos(u)/\sin^2 u$ into $\csc u \cot u.$)

Answer 3

$\int \frac{\arcsin(x)}{x^2}dx=\int \arcsin(x) d(-\frac{1}{x})=-\frac{1}{x}\arcsin(x)+\int \frac{dx}{x \sqrt{1-x^2}}= -\frac{1}{x}\arcsin(x)+\int \frac{dx}{x^2 \sqrt{\frac{1}{x^2}-1}}=-\frac{1}{x}\arcsin(x)+\int \frac{d(-\frac{1}{x})}{\sqrt{\frac{1}{x^2}-1}}=-\frac{1}{x}\arcsin(x)-\int \frac{dy}{\sqrt{y^2-1}}=-\frac{1}{x}\arcsin(x)-\ln(y+\sqrt{y^2-1})=-\frac{1}{x}\arcsin(x)-\ln(\frac{1}{x}+\sqrt{(1/x)^2-1})$

Answer 4

Consider $$ \int\frac{\arcsin(x)}{x^2}dx. $$ We may use integration by parts where \begin{align*} u&=\arcsin(x)&dv&=\frac{dx}{x^2}\\ du&=\frac{dx}{\sqrt{1-x^2}}&v&=-\frac{1}{x}. \end{align*} In this case, $$ \int\frac{\arcsin(x)}{x^2}dx=-\frac{\arcsin(x)}{x}-\int-\frac{dx}{x\sqrt{1-x^2}}=-\frac{\arcsin(x)}{x}+\int\frac{dx}{x\sqrt{1-x^2}}. $$ Focusing on the last term, we would like to use $u$-substitution, $u^2=1-x^2$ so that $2udu=-2xdx$. In other words, $dx=-\frac{udu}{x}$. Note that in this substitution, it only makes sense when $1-x^2\geq 0$, i.e., $|x|\leq 1$ and we may assume that $u$ is positive. Using this substitution, we get the integral $$ \int\frac{1}{x\sqrt{u^2}}\cdot\frac{-udu}{x}=-\int\frac{u}{|u|}\frac{du}{x^2}=-\int\frac{du}{1-u^2}=-\int\frac{du}{1-u^2}. $$ We can now use partial fractions decomposition noting that $$ \frac{1}{1-u^2}=\frac{1}{2(1-u)}+\frac{1}{2(1+u)} $$ This ends up being $$ -\left(-\frac{1}{2}\ln|1-u|+\frac{1}{2}\ln|1+u|\right). $$ Now, finally, combine everything and you have your integral (near $x=0$). After substitution, one gets $$ -\frac{\arcsin(x)}{x}+\frac{1}{2}\ln\left|1-\sqrt{1-x^2}\right|-\frac{1}{2}\ln\left|1+\sqrt{1-x^2}\right| $$ whose derivative is $\frac{\arcsin(x)}{x^2}$ (checked with Maple).

Comparing this with the accepted answer, we can use log rules to get $$ -\frac{\arcsin(x)}{x}+\frac{1}{2}\ln\left|1-\sqrt{1-x^2}\right|-\frac{1}{2}\ln\left|1+\sqrt{1-x^2}\right|=-\frac{\arcsin(x)}{x}+\frac{1}{2}\ln\left|\frac{1-\sqrt{1-x^2}}{1+\sqrt{1-x^2}}\right|. $$ Multiplying the numerator and denominator of the argument to the logarithm by the conjugate of the numerator, we get $$ -\frac{\arcsin(x)}{x}+\frac{1}{2}\ln\left|\frac{1-(1-x^2)}{(1+\sqrt{1-x^2})^2}\right|=-\frac{\arcsin(x)}{x}+\frac{1}{2}\ln\left|\frac{x}{1+\sqrt{1-x^2}}\right|^2. $$ Using log rules the $2$'s cancel and we can break up the logarithm, leaving $$ -\frac{\arcsin(x)}{x}+\ln\left|\frac{x}{1+\sqrt{1-x^2}}\right|=-\frac{\arcsin(x)}{x}-\ln\left|1+\sqrt{1-x^2}\right|+\ln\left|x\right| $$ Finally, remember $+C$.