Question

If $xf\left(x\right)=3f^2\left(x\right)+2,$ then $\int \frac{2x^2-12xf\left(x\right)+f\left(x\right)}{\left(6f\right(x)-x)(x^2-f(x){)}^2}dx$ equals

• A. $\frac{1}{x^2-f\left(x\right)}+c$
• B. $\frac{1}{x^2+f\left(x\right)}+c$
• C. $\frac{1}{x-f\left(x\right)}+c$
• D. $\frac{1}{x+f\left(x\right)}+c$

Answer 1

The correct option is A $\frac{1}{x^2-f\left(x\right)}+c$

Given $xf\left(x\right)=3f^2\left(x\right)+2$
$\Rightarrow x{f}^{\prime }\left(x\right)+f\left(x\right)=6f\left(x\right)f^{\prime }\left(x\right)$
$\Rightarrow f^{\prime }\left(x\right)=\frac{f\left(x\right)}{6f\left(x\right)-x}$
$\Rightarrow f\left(x\right)=f^{\prime }\left(x\right)\left[6f\right(x)-x]$
Now, consider $I=\int \frac{2x^2-12xf\left(x\right)+f\left(x\right)}{\left(6f\right(x)-x)(x^2-f(x){)}^2}dx$
$I=\int \frac{2x(x-6f(x\left)\right)+f\left(x\right)}{\left(6f\right(x)-x)(x^2-f(x){)}^2}dx$
$=-\int \frac{2x-f^{\prime }\left(x\right)}{(x^2-f(x){)}^2}dx$
Put $x^2-f\left(x\right)=t$
$\Rightarrow 2x-f^{\prime }\left(x\right)=\frac{dt}{dx}$
$\Rightarrow [2x-f^{\prime }(x\left)\right]dx=dt$
So, $I=-\int \frac{dt}{t^2}$
$\Rightarrow I=\frac{1}{t}+C$
$I=\frac{1}{x^2-f\left(x\right)}+C$