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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?

  • 0
    You know you could select an answer as accepted answer. (Or vote for answers you like)2012-03-15

2 Answers 2

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

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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    $\int \frac{1}{(1-kx^2)^{\frac{3}{2}}} = \frac{x}{1-kx^2}$ where k > 0 here2012-03-12