Question

If $I=\int \frac{dx}{{\left(2ax+x^2\right)}^{\frac{3}{2}}}$, then $I$ is equal to

• A. $-\frac{x+a}{\sqrt{2ax+x^2}}+c$
• B. $-\frac{1}{a}\frac{x+a}{\sqrt{2ax+x^2}}+c$
• C. $-\frac{1}{a^2}\frac{x+a}{\sqrt{2ax+x^2}}+c$
• D. $-\frac{1}{a^3}\frac{x+a}{\sqrt{2ax+x^3}}+c$

Answer 1

The correct option is C $-\frac{1}{a^2}\frac{x+a}{\sqrt{2ax+x^2}}+c$

We have,

$I=\int \frac{dx}{{(2ax+x^2)}^{3/2}}$

$I=\int \frac{dx}{{({(x+a)}^2-a^2)}^{3/2}}$

Let $t=x+a$

$dt=dx$

$I=\int \frac{dt}{{(t^2-a^2)}^{3/2}}$

Put $t=a\sec \left(p\right)$

$\frac{dt}{dp}=a\sec \left(p\right)\tan \left(p\right)$

$dt=a\sec \left(p\right)\tan \left(p\right)dp$

Therefore,

$I=\int \frac{a\sec \left(p\right)\tan \left(p\right)}{{(a^2{\sec }^{2}p-a^2)}^{3/2}}dp$

$I=\int \frac{a\sec \left(p\right)\tan \left(p\right)}{a^3{({\sec }^{2}p-1)}^{3/2}}dp$

$I=\frac{1}{a^2}\int \frac{\sec \left(p\right)\tan \left(p\right)}{{\left({\tan }^{2}p\right)}^{3/2}}dp$

$I=\frac{1}{a^2}\int \frac{\sec \left(p\right)}{\left({\tan }^{2}p\right)}dp$

$I=\frac{1}{a^2}\int \frac{\frac{1}{\cos p}}{\frac{{\sin }^{2}p}{{\cos }^{2}p}}dp$

$I=\frac{1}{a^2}\int \frac{\cos p}{{\sin }^{2}p}dp$

Let $m=\sin p$

$dm=\cos pdp$

Therefore,

$I=\frac{1}{a^2}\int \frac{1}{m^2}dm$

$I=\frac{1}{a^2}(-\frac{1}{m})+C$

On putting all values, we get

$I=-\frac{1}{a^2}\left(\frac{1}{\sin p}\right)+C$

$I=-\frac{1}{a^2}\left(\frac{1}{\sin \left({\sec }^{-1}\left(\frac{t}{a}\right)\right)}\right)+C$

$I=-\frac{1}{a^2}\left(\frac{\frac{1}{a\sqrt{\frac{t^2}{a^2}-1}}}{t}\right)+C$

$I=-\frac{1}{a^3}\left(\frac{t}{\sqrt{\frac{t^2}{a^2}-1}}\right)+C$

$I=-\frac{1}{a^3}\left(\frac{x+a}{\sqrt{\frac{{(x+a)}^2}{a^2}-1}}\right)+C$

$I=-\frac{1}{a^2}\left(\frac{x+a}{\sqrt{{(x+a)}^2-a^2}}\right)+C$

$I=-\frac{1}{a^2}\left(\frac{x+a}{\sqrt{x^2+2ax}}\right)+C$

Hence, this is the answer.