Can anybody help me find the value of this integral? I utterly have no clue on how to solve this and $r$ is a constant.
$$\int \frac{dx}{\left({x^2+r^2}\right)^{3/2}}$$
The answer is given as $\dfrac{x}{{\left(x^2+r^2\right)}^{1/2}}$
I even tried differentiating the answer, but didn't get the original function back.
You can compute that integral doing the substitution $x=r\tan\theta$ and $\mathrm dx=r(1+\tan^2\theta)\,\mathrm d\theta$. The correct answer is $\dfrac x{r^2(x^2+r^2)^{\frac12}}$.
I get \begin{align} \frac d{dx}\frac x{(x^2+r^2)^{1/2}} &=\frac1{(x^2+r^2)^{1/2}}-\frac{x^2}{(x^2+r^2)^{3/2}}\\ &=\frac{x^2+r^2-x^2}{(x^2+r^2)^{3/2}}=\frac{r^2}{(x^2+r^2)^{3/2}} \end{align} which is your original integrand, save for a constant factor $r^2$.
Systematic ways to attack this include the substitutions $x=r\tan t$ or $x=r\sinh y$.
It's simpler if you first get rid of $r$: substitute $x=rt$, so your integral becomes $$ \int\frac{1}{r^3(t^2+1)^{3/2}}r\,dt $$ We can ignore $1/r^2$ for the time being and reinsert it at the end: now $$ \int\frac{1}{(t^2+1)^{3/2}}\,dt= \int\frac{t^2+1-t^2}{(t^2+1)^{3/2}}\,dt= \int\frac{1}{(t^2+1)^{1/2}}\,dt+ \int t\frac{-t}{(t^2+1)^{3/2}}\,dt $$ Do the last by parts as indicated: $$ \int t\frac{-t}{(t^2+1)^{3/2}}\,dt= t\frac{1}{(t^2+1)^{1/2}}-\int\frac{1}{(t^2+1)^{1/2}}\,dt $$ The two integrals now cancel! $$ \int\frac{1}{(t^2+1)^{3/2}}\,dt=\frac{t}{(t^2+1)^{1/2}}+c $$ Reinsert the factor $1/r^2$ and back substitute: $$ \int\frac{1}{(x^2+r^2)^{3/2}}\,dx= \frac{1}{r^2}\frac{x/r}{(x^2/r^2+1)^{1/2}}+c=\frac{1}{r^2}\frac{x}{(x^2+r^2)^{1/2}}+c $$
Set $x=r\tan\theta\implies \mathrm dx=r\sec^2\theta\mathrm d\theta$. $$\begin{aligned}\int\dfrac{1}{(x^2+r^2)^{3/2}}\mathrm dx&\mapsto \int\dfrac{r\sec^2\theta}{r^3\sec^3\theta}\mathrm d\theta\\&=\dfrac{1}{r^2}\int\cos\theta\mathrm d\theta\\&=\dfrac{1}{r^2}\sin\theta=\dfrac{1}{r^2}\cdot\dfrac{x}{\sqrt{x^2+r^2}}+C\end{aligned}$$
