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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?

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Plug x = tan{ y / (k^1/2) }, and you will get, (after simplification) integ {cos(y) / (k^1/2) } from 1 to -1 – Tomarinator Mar 12 '12 at 8:31
sorry i meant plug x = {y} / k^1/2 – Tomarinator Mar 12 '12 at 8:44
You know you could select an answer as accepted answer. (Or vote for answers you like) – Jeremy Carlos Mar 15 '12 at 0:08

$\int \dfrac{1}{(1+kx^2)^{3/2}}dx$

Put $1+kx^2=t$, Then, $$2kx\cdot dx = dt \text{. Also, } x = \sqrt{t-1\over k}$$ $$2kx\cdot dx = dt$$

$$dx = \frac{dt}{2kx} = \frac{\sqrt{k}dt}{2k\sqrt{t-1}}= \frac{dt}{2\sqrt{k}\sqrt{t-1}}$$

$\int \dfrac{1}{(1+kx^2)^{3/2}}dx=$ $\int \dfrac{1}{2t^{1.5}\sqrt{k}\sqrt{t-1}}dt$ $$\dfrac{1}{2\sqrt{k}}\int \dfrac{1}{t^{1.5}.\sqrt{t-1}}dt=\dfrac{1}{2\sqrt{k}}\dfrac{2\sqrt{t-1}}{\sqrt{t}}=\dfrac{1}{\sqrt{k}}\dfrac{\sqrt{t-1}}{\sqrt{t}}$$

$$=\dfrac{1}{\sqrt{k}}\dfrac{\sqrt{t-1}}{\sqrt{t}}$$ $$=\dfrac{1}{\sqrt{k}}\dfrac{\sqrt{kx^2}}{\sqrt{1+kx^2}}$$ $$=\dfrac{x}{\sqrt{1+kx^2}}$$

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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

$$ \frac{1}{2\sqrt{k}} \int 2 du = \frac{1}{2\sqrt{k}} 2u = \frac{u}{\sqrt{k}} = \left(\sqrt{\frac{t-1}{tk}}\right) = \frac{x}{\sqrt{1+kx^2}}$$

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So what if $k<0$? – Norbert Mar 12 '12 at 16:35
$\int \frac{1}{(1-kx^2)^{\frac{3}{2}}} = \frac{x}{1-kx^2}$ where $k > 0$ here – Kirthi Raman Mar 12 '12 at 18:48

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