How do you find the indefinite integral for $\int \frac1{(x^2+1)^{3/2}} dx $?

Question

How do you find the indefinite integral for $\int \frac1{(x^2+1)^{3/2}} dx $ ?

According to the Apostol's Calculus Vol1 textbook this can be done by integration by substitution but not sure how.

Answer 1

Substitute $x=\tan{u}\implies dx=\sec^2{u}\,du$.

$$\therefore\int\frac{1}{(x^2+1)^{3/2}}\,dx=\int\frac{\sec^2{u}}{\left(\tan^2{u}+1\right)^{3/2}}\,du=\int\frac{\sec^2{u}}{\sec^3{u}}\,du=\int\cos{u}\,du=\sin{u}+C$$ Now, note that $u=\tan^{-1}x$. Therefore, $$\int\frac{1}{(x^2+1)^{3/2}}\,dx=\sin(\tan^{-1}x)+C$$

As @martycohen mentioned below, one can simplify to $$\sin(\tan^{-1}x)+C=\frac{x}{\sqrt{1+x^2}}+C$$

Answer 2

HINT: Differentiate the function $$\dfrac{x}{\sqrt{x^2+1}}$$

Answer 3

Another possible approach.

Considering $$I=\int\frac{{dx}}{\left(x^2+1\right)^{3/2}}$$ change variable $x=\sinh(y)$, $dx=\cosh(y)\,dy$ and get $$I=\int\text{sech}^2(y)\,dy=\tanh (y)+C$$ Now, back to $x$, $$\tanh (y)=\frac{\sinh(y)}{\cosh(y)}=\frac x {\sqrt{1+x^2}}$$