Question

Solve this:

Spamming accounts will be reported.
​

Answer 1

Topic :-

Indefinite Integration

To Integrate :-

$\displaystyle \int\dfrac{dx}{(2ax+x^2)^{3/2}}$

Solution :-

$\displaystyle \int\dfrac{dx}{(2ax+x^2)^{3/2}}$

Do manipulation in denominator,

$\displaystyle \int\dfrac{dx}{(x^2+2ax+a^2-a^2)^{3/2}}$

$\displaystyle \int\dfrac{dx}{((x+a)^2-a^2)^{3/2}}$

$\bold{Substitute\:\: x + a = a\sec\theta,}$

$\sec\theta=\left(\dfrac{x+a}{a}\right)$

$dx=a\sec\theta\tan\theta \:d\theta$

$\displaystyle \int\dfrac{a\sec\theta\tan\theta}{(a^2\sec^2\theta-a^2)^{3/2}}\:d\theta$

$\displaystyle \int\dfrac{a\sec\theta\tan\theta}{(a^2(\sec^2\theta-1))^{3/2}}\:d\theta$

$\displaystyle \int\dfrac{a\sec\theta\tan\theta}{(a^2\tan^2\theta)^{3/2}}\:d\theta$

$(\because \sec^2\theta-1=\tan^2\theta)$

$\displaystyle \int\dfrac{a\sec\theta\tan\theta}{(a\tan\theta)^{3}}\:d\theta=\int\dfrac{a\sec\theta\tan\theta}{a^3\tan^3\theta}\:d\theta$

$\displaystyle \int\dfrac{\sec\theta}{a^2\tan^2\theta}\:d\theta$

$\displaystyle \dfrac{1}{a^2}\int\dfrac{\cos^2\theta}{\cos\theta\sin^2\theta}\:d\theta$

$\displaystyle \dfrac{1}{a^2}\int\dfrac{\cos\theta}{\sin^2\theta}\:d\theta$

$\bold{Substitute\:\: t = \sin\theta,}$

$dt=\cos\theta\:d\theta$

$\displaystyle \dfrac{1}{a^2}\int\dfrac{1}{t^2}\:dt$

$\displaystyle \dfrac{1}{a^2}\left(\dfrac{t^{-2+1}}{-2+1}\right)+C$

$\left(\because \displaystyle \int z^n\:dz=\dfrac{z^{n+1}}{n+1}+C, where\:C\:is\:constant\:of\:integration.\right)$

$\displaystyle -\dfrac{1}{a^2}\left(\dfrac{1}{t}\right)+C$

Put back value of 't',

$\displaystyle -\dfrac{1}{a^2}\left(\dfrac{1}{\sin\theta}\right)+C$

Put back value of sinθ,

$\sin\theta=\sqrt{1-\dfrac{1}{\sec^2\theta}}$

$\sin\theta=\sqrt{1-\left(\dfrac{a}{x+a}\right)^2}$

$\sin\theta=\sqrt{\dfrac{(x+a)^2-a^2}{(x+a)^2}}$

$\sin\theta=\dfrac{\sqrt{2ax+x^2}}{x+a}$

Putting back,

$\displaystyle -\dfrac{1}{a^2}\left(\dfrac{x+a}{\sqrt{2ax+x^2}}\right)+C$

Answer :-

$\underline{\boxed{\displaystyle \int\dfrac{dx}{(2ax+x^2)^{3/2}}=\displaystyle -\dfrac{1}{a^2}\left(\dfrac{x+a}{\sqrt{2ax+x^2}}\right)+C}}$