Question

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

Answer 1

Suppose $g ( x )=\int \frac{f( x ) dx }{ x ^2( x +1)^3}$.....(1)
$\left[12^{\text {th }} \& 13^{\text {th }} 7-01-2007\right]$
$=\int\left(\frac{ A }{ x }+\frac{ B }{ x ^2}+\frac{ C }{ x +1}+\frac{ D }{( x +1)^2}+\frac{ E }{( x +1)^3}\right) dx = A \ln x -\frac{ B }{ x }+ C \ln (1+ x )-\frac{ D }{1+ x }-\frac{ E }{2( x +1)^2}$
since $g(x)$ is a rational function hence logarithmic functions must not be there $\Rightarrow A=C=0$
$g(x)=\int\left(\frac{B}{x^2}+\frac{D}{(x+1)^2}+\frac{E}{(x+1)^3}\right) d x$.....(2)
comparing $N ^{ r}$ of $(1)$ and (2)
$f(x)=B(x+1)^3+D x^2(x+1)+E x^2$
$f(x)=(B+D) x^3+(3 B+D+E) x^2+3 B x+B$
$\therefore f ( x )$ is quadratic function, hence $B + D =0$
$\text { also } f(0)=1 \text { gives } B=1 \Rightarrow D=-1$
$\therefore f ( x )=(2+ E ) x ^2+3 x +1$
$f^{\prime}(x)=2(2+E) x+3$
$f^{\prime}(0)=3$