Question

If $I=\int \frac{x^2}{(x-a)(x-b)}dx$, then I equals

• A. $x+\frac{1}{a-b}\log \left|\frac{x-a}{x-b}\right|+C$
• B. $x+\frac{1}{a-b}\log {\left|\frac{x-a}{x-b}\right|}^{a^2+b^2}+C$
• C. $x+\frac{1}{a-b}\{a^2\log |x-a|-b^2\log |x-b|\}+C$
• D. none of these

Answer 1

The correct option is C $x+\frac{1}{a-b}\{a^2\log |x-a|-b^2\log |x-b|\}+C$

$I=\int \frac{x^2}{(x-a)(x-b)}dx=\int (\frac{a^2}{(x-b)(x-a)}-\frac{b^2}{(a-b)(x-b)}+1)dx$
$=\frac{-b^2}{a-b}\int \frac{1}{x-b}dx+\frac{a^2}{a-b}\int \frac{1}{x-a}dx+\int dx$
$=\frac{-b^2}{a-b}\log (x-b)+\frac{a^2}{a-b}\log (x-a)+x$
$=x+\frac{1}{a-b}\left(a^2\log (x-a)-b^2\log (x-b)\right)+c$