The value of $\mathrm{I}=\int \frac{x^2}{(\mathrm{a}+\mathrm{b} x)^2} \mathrm{~d} x$ is
The value of $\mathrm{I}=\int \frac{x^2}{(\mathrm{a}+\mathrm{b} x)^2} \mathrm{~d} x$ is
$\frac{1}{b^3}\left[a+b x+2 a \log |a+b x|-\frac{a^2}{a+b x}\right]+c$,
(where c is the constant of integration)
$\frac{1}{b^3}\left[a+b x-2 a \log |a+b x|+\frac{a^2}{a+b x}\right]+c$,
(where c is the constant of integration)
$\frac{1}{b^3}\left[a+b x-2 a \log |a+b x|-\frac{a^2}{a+b x}\right]+c$, (where c is the constant of integration)
$\frac{1}{b^3}\left[a+b x+2 a \log |a+b x|+\frac{a^2}{a+b x}\right]+c$,
(where c is the constant of integration)
Solution
$\begin{array}{rl}\quad I & =\int \frac{x^2}{(a+b x)^2} d x \\ & \text { Let } a+b x=t \Rightarrow x=\frac{t-a}{b} \\ \therefore \quad b & d x=d t \\ \therefore \quad & =\frac{d t}{b} \\ \therefore \quad I & =\int \frac{\left(\frac{t-a}{b}\right)^2}{t^2} \times \frac{d t}{b} \\ & =\frac{1}{b^3} \int \frac{(t-a)^2}{t^2} d t \\ & =\frac{1}{b^3} \int \frac{t^2-2 a t+a^2}{t^2} d t \\ & =\frac{1}{b^3}\left[\int 1 d t-2 a \int \frac{1}{t}+a^2 \int \frac{1}{t^2} d t\right]\end{array}$
$\begin{aligned} & =\frac{1}{b^3}\left[t-2 a \log |t|-\frac{a^2}{t}\right]+c \\ & =\frac{1}{b^3}\left[a+b x-2 a \log |a+b x|-\frac{a^2}{a+b x}\right]+c\end{aligned}$