How to integrate $\int\frac{dx}{(A^2+x^2)^{3/2}}$

Question

I have the following to integrate:

$$\int\frac{dx}{(A^2+x^2)^{3/2}}$$

Where, $A$ is a constant.

Not really sure what technique to use. Any hint would be helpful.

Answer 1

let $x=A\tan\theta\implies dx=A\sec^2\theta\ d\theta$ $$\int \frac{1}{(A^2+x^2)^{3/2}}dx$$ $$=\int \frac{1}{(A^2+A^2\tan^2\theta)^{3/2}}(A\sec^2\theta\ d\theta)$$ $$=\int \frac{A\sec^2\theta\ d\theta}{A^{3}\sec^3\theta}$$ $$=\frac{1}{A^2}\int \cos \theta\ d\theta$$ $$=\frac{1}{A^2}\sin\theta+C$$ $$=\frac{1}{A^2}\sin\left(\tan^{-1}\frac{x}{A}\right)+C$$ $$=\frac{1}{A^2}\sin\left(\sin^{-1}\frac{x/A}{\sqrt{1+(x/A)^2}}\right)+C$$ $$=\color{red}{\frac{x}{A^2\sqrt{A^2+x^2}}+C}$$

Answer 2

Hint: Use the trigonometric substitution $x=A\tan u$ and $dx=A \sec^2 u$.

Answer 3

Use substitution : $x=\tan t$. You get the integral $$\displaystyle\int\frac{\mathrm d\mkern1.5mu t}{A^2(1+\tan^2t)^2}=\int \cos t\,\mathrm d\mkern1.5mu t=\frac1{A^2}\sin t=\frac1{A^2}\sin(\arctan t)=\frac{x}{A^2\sqrt{1+x^2}}.$$

Answer 4

$$\int\frac{dx}{(A^2+x^2)^{3/2}}$$

Hint:

Substitute $x=A\tan(t)$ and $dx=A\sec^2(t)dt$. Then $(x^2+A^2)^{3/2}=(A^2\tan^2(t)+A^2)^{3/2}=A^3\sec^3(t)$ and $t=\arctan(\frac x A)$

Full answer:

$$A\int\frac{\cos t}{A^3}dt=\frac{1}{A^2}\int \cos t dt=\frac{\sin t}{A^2}+\mathcal C=\frac{1}{A^2}\sin\left(\arctan(x/A)\right)+\mathcal C=\color{red}{\frac{x}{A^2\sqrt{x^2+A^2}}+\mathcal C}$$

Answer 5

The substitution $$u=\frac{1}{(a^2+x^2)^\frac{1}{2}}$$ transforms the integral to $$-\frac{1}{a^2}\int \frac{au}{\sqrt{1-a^2u^2}}d(au)=\frac{\sqrt{1-(au)^2}}{a^2}$$ Upon resubstitution the last term is seen to be same as $$\frac{x}{a^2\sqrt{a^2+x^2}}$$