Question

Let $f\left(x\right)$ be a quadratic function such that $f\left(0\right)=1$ and $\int \frac{f\left(x\right)}{x^2(x+1{)}^3}dx$ is a rational function. Then the value of $f^{\prime }\left(0\right)$ is

Answer 1

Suppose $g\left(x\right)=\int \frac{f\left(x\right)}{x^2(x+1{)}^3}dx\cdots \left(1\right)$
$g\left(x\right)=\int (\frac{A}{x}+\frac{B}{x^2}+\frac{C}{x+1}+\frac{D}{(x+1{)}^2}+\frac{E}{(x+1{)}^3})dx$
$=A\ln x-\frac{B}{x}+C\ln (1+x)-\frac{D}{1+x}-\frac{E}{2(x+1{)}^2}+F,$ where $F$ is an integration constant.
Since $g\left(x\right)$ is a rational function, hence logarithmic functions must not be there.
$\Rightarrow A=C=0$
$g\left(x\right)=\int (\frac{B}{x^2}+\frac{D}{(x+1{)}^2}+\frac{E}{(x+1{)}^3})dx\cdots \left(2\right)$

Comparing numerator of $\left(1\right)$ and $\left(2\right),$
$f\left(x\right)=B(x+1{)}^3+D{x}^2(x+1)+E{x}^2$
$\Rightarrow f\left(x\right)=(B+D)x^3+(3B+D+E)x^2+3Bx+B$
Since $f\left(x\right)$ is a quadratic equation, hence $B+D=0$
Also, $f\left(0\right)=1$ gives $B=1$
$\Rightarrow D=-1$
$\therefore f\left(x\right)=(2+E)x^2+3x+1$
$f^{\prime }\left(x\right)=2(2+E)x+3$
$f^{\prime }\left(0\right)=3$