How to evaluate $\displaystyle\int {1\over (1+kx^2)^{3/2}}dx$ What change of variable should I use to integrate $$\displaystyle\int {1\over (1+kx^2)^{3/2}}dx$$
I know the answer is $$\displaystyle x\over \sqrt{kx^2+1}.$$ Maybe a trig or hyperbolic function?
 A: $\int \dfrac{1}{(1+kx^2)^{3/2}}dx$
Put $1+kx^2=t$, Then, 
$$2kx\cdot dx = dt \text{. Also, } x = \sqrt{t-1\over k}$$
$$2kx\cdot dx = dt$$
$$dx = \frac{dt}{2kx} = \frac{\sqrt{k}dt}{2k\sqrt{t-1}}= \frac{dt}{2\sqrt{k}\sqrt{t-1}}$$
$\int \dfrac{1}{(1+kx^2)^{3/2}}dx=$ $\int \dfrac{1}{2t^{1.5}\sqrt{k}\sqrt{t-1}}dt$
$$\dfrac{1}{2\sqrt{k}}\int \dfrac{1}{t^{1.5}.\sqrt{t-1}}dt=\dfrac{1}{2\sqrt{k}}\dfrac{2\sqrt{t-1}}{\sqrt{t}}=\dfrac{1}{\sqrt{k}}\dfrac{\sqrt{t-1}}{\sqrt{t}}$$
$$=\dfrac{1}{\sqrt{k}}\dfrac{\sqrt{t-1}}{\sqrt{t}}$$
$$=\dfrac{1}{\sqrt{k}}\dfrac{\sqrt{kx^2}}{\sqrt{1+kx^2}}$$
$$=\dfrac{x}{\sqrt{1+kx^2}}$$
A: For clarity purposes 
$$
\begin{align*}
\dfrac{1}{2\sqrt{k}}\int \dfrac{1}{t^{\frac{3}{2}} \sqrt{t-1}}dt &= \dfrac{1}{2\sqrt{k}}\int \dfrac{\sqrt{t}}{t^{2} \sqrt{t-1}}dt\\
&= \dfrac{1}{2\sqrt{k}}\int \dfrac{1}{t^{2} \sqrt{1-\frac{1}{t}}}dt \tag{A}
\end{align*}
$$
Now in this integral above substitute
$$ \sqrt{1-\frac{1}{t}} = u$$ which would imply
$$ \frac{1}{2t^2 \sqrt{(1-\frac{1}{t})}} dt = du$$
Thus, $(A)$ would simplify to 
$$ \frac{1}{2\sqrt{k}} \int 2 du = \frac{1}{2\sqrt{k}} 2u = \frac{u}{\sqrt{k}}  = \left(\sqrt{\frac{t-1}{tk}}\right) = \frac{x}{\sqrt{1+kx^2}}$$ 
