Question

The value of $\int \frac{x^2}{(a+bx{)}^2}dx$ is:
(where $a\ne b$ and $C$ is constant of integration)

• A. $\frac{1}{b^3}[a+bx+2a\ln |a+bx|+\frac{a^2}{a+bx}]+C$
• B. $\frac{1}{a^3}[a+bx+2a\ln |a+bx|+\frac{a^2}{a+bx}]+C$
• C. $\frac{1}{b^3}[a+bx-2a\ln |a+bx|-\frac{a^2}{a+bx}]+C$
• D. $\frac{1}{a^3}[a+bx-2a\ln |a+bx|-\frac{a^2}{a+bx}]+C$

Answer 1

The correct option is C $\frac{1}{b^3}[a+bx-2a\ln |a+bx|-\frac{a^2}{a+bx}]+C$

Let $I=\int \frac{x^{2}dx}{(a+bx{)}^2}$
Let $a+bx=t$
$\Rightarrow x=\left(\frac{t-a}{b}\right)$
$\Rightarrow dx=\frac{dt}{b}$
$\therefore I=\frac{1}{b^3}\int \frac{t^2-2at+a^2}{t^2}dt$
$\Rightarrow I=\frac{1}{b^3}\int (1-\frac{2a}{t}+\frac{a^2}{t^2})dt$
$\Rightarrow I=\frac{1}{b^3}[t-2a\ln |t|-\frac{a^2}{t}]+C$
$\therefore I=\frac{1}{b^3}[a+bx-2a\ln |a+bx|-\frac{a^2}{a+bx}]+C$