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

Answer: 3

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 $\Rightarrow A=C=0$
$g(x)=\int \left(\frac{B}{x^2}+\frac{D}{(x+1{)}^2}+\frac{E}{(x+1{)}^3}\right)dx$.....(2)
comparing $N^r$ of $(1)$ and (2)
$f(x)=B(x+1{)}^3+D{x}^2(x+1)+E{x}^2$
$f(x)=(B+D)x^3+(3B+D+E)x^2+3Bx+B$
$\therefore f(x)$ is quadratic function, hence $B+D=0$
$\text{ also }f(0)=1\text{ gives }B=1\Rightarrow D=-1$
$\therefore f(x)=(2+E)x^2+3x+1$
$f^{\prime }(x)=2(2+E)x+3$
$f^{\prime }(0)=3$