∫ x2(ax + b)-2 dx is equal to

Question

∫ x2(ax + b)-2 dx is equal to (A) (2/a2)(x -(b/a) log(ax+b))+ C (B) (2/a2)(x -(b/a) log(ax+b)) -(x2/a(ax+b))+ C (C) (2/a2)(x +(b/a) log(ax+b)) +(x2/a(ax+b))+ C (D) (2/a2)(x +(b/a) log(ax+b)) -(x2/a(ax+b))+ C

Tardigrade Admin · Accepted Answer

I= ∫(x2/(ax+b)2) dx Put ax +b =t ⇒ dx =(1/a) dt ∴ I =(1/a3)∫((t-b)2/t2) dt =(1/a3) ∫(1+(b2/t2)-(2b/t)) dt = (1/a3) (t -(b2/t)-2blogt) + C (1/a3)(ax + b -(b2/ax+b)-2b log(ax+b)) + C = (2/a2) (x-(b/a) log(ax+b))-(x2/a(ax+b)) + C

Q. $\int x^{2} (a x + b)^{- 2} d x$ is equal to

$\frac{2}{a ^{2}} (x - \frac{b}{a} l o g (a x + b)) + C$

$\frac{2}{a ^{2}} (x - \frac{b}{a} l o g (a x + b)) - \frac{x ^{2}}{a ( a x + b )} + C$

$\frac{2}{a ^{2}} (x + \frac{b}{a} l o g (a x + b)) + \frac{x ^{2}}{a ( a x + b )} + C$

$\frac{2}{a ^{2}} (x + \frac{b}{a} l o g (a x + b)) - \frac{x ^{2}}{a ( a x + b )} + C$

Q. ∫x2(ax+b)−2dx is equal to

a22​(x−ab​log(ax+b))+C

a22​(x−ab​log(ax+b))−a(ax+b)x2​+C

a22​(x+ab​log(ax+b))+a(ax+b)x2​+C

a22​(x+ab​log(ax+b))−a(ax+b)x2​+C

I=∫(ax+b)2x2​dx Put ax+b=t⇒dx=a1​dt ∴I=a31​∫t2(t−b)2​dt=a31​∫(1+t2b2​−t2b​)dt =a31​(t−tb2​−2blogt)+C a31​(ax+b−ax+bb2​−2blog(ax+b))+C =a22​(x−ab​log(ax+b))−a(ax+b)x2​+C

Q. $\int x^{2} (a x + b)^{- 2} d x$ is equal to

$\frac{2}{a ^{2}} (x - \frac{b}{a} l o g (a x + b)) + C$

$\frac{2}{a ^{2}} (x - \frac{b}{a} l o g (a x + b)) - \frac{x ^{2}}{a ( a x + b )} + C$

$\frac{2}{a ^{2}} (x + \frac{b}{a} l o g (a x + b)) + \frac{x ^{2}}{a ( a x + b )} + C$

$\frac{2}{a ^{2}} (x + \frac{b}{a} l o g (a x + b)) - \frac{x ^{2}}{a ( a x + b )} + C$