Question

Let $f\left(x\right)$ be a polynomial of degree three such that $f\left(0\right)=1,f\left(1\right)=2$ and $f\left(x\right)$ has a critical point at $x=0$ where $f\left(x\right)$ does not have a local extremum, then $\int \frac{f\left(x\right)}{x^2+1}dx$ is equal to

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

Answer 1

Answer: (D)

$\frac{1}{2}(x^2-\log (x^2+1\left)\right)+{\tan }^{-1}x+c$

Given that $f\left(x\right)$ is a polynomial of degree 3.
Let $f\left(x\right)=a{x}^3+b{x}^2+cx+d$.
Given that $f\left(0\right)=1$
$\Rightarrow d=1$
Also given that $f\left(x\right)$ has a critical point at $x=0$ where it does not have a local extremum.
$\Rightarrow f^{\prime }\left(x\right)=0$ and $f^{\prime \prime }\left(x\right)=0$ at $x=0$....(1)
$f^{\prime }\left(x\right)=3a{x}^2+2bx+c=0$, $f^{\prime \prime }\left(x\right)=6ax+2b$....(2)
Using (1) and (2), we get $c=0;b=0$
$\Rightarrow f\left(x\right)=a{x}^3+1$
Also given that $f\left(1\right)=2\Rightarrow a+1=2\Rightarrow a=1$
$\Rightarrow f\left(x\right)=x^3+1$
So, $\int \frac{x^3+1}{x^2+1}dx.=\int \frac{x(x^2+1)+(1-x)}{x^2+1}=\int [x+\frac{1-x}{1+x^2}]dx$
$=\int x+\frac{1}{1+x^2}-\frac{1}{2}\frac{2x}{1+x^2}.dx$
$=\frac{x^2}{2}+ta{n}^{-1}x-\frac{1}{2}log(1+x^2)+C$
$=\frac{1}{2}[x^2-log(1+x^2\left)\right]+ta{n}^{-1}x+C$