Question

Let $p\left(x\right)$ be a polynomial such that $\frac{p(x+1)}{p\left(x\right)}=\frac{x^2+x+1}{x^2-x+1}$ and $p\left(2\right)=3$, then ${\int }_0^1{\tan }^{-1}\left(p\right(x\left)\right)\cdot {\tan }^{-1}\sqrt{\frac{x}{1-x}}dx$ is

• A. $\frac{\pi }{2}\ln 2$
• B. $\frac{\pi }{4}\ln 2$
• C. $\frac{\pi }{8}\ln 8$
• D. $\frac{\pi }{2}\ln 4$

Answer 1

The correct option is B $\frac{\pi }{4}\ln 2$

$p\left(x\right)(x^2+x+1)=(x^2-x+1)p(x+1)$
$\frac{p\left(3\right)}{p\left(2\right)}\times \frac{p\left(4\right)}{p\left(3\right)}\times \cdots \times \frac{p(x+1)}{p\left(x\right)}=\frac{7}{3}\times \frac{13}{7}\times \cdots \times \frac{x^2+x+1}{x^2-x+1}$
$\frac{p(x+1)}{p\left(2\right)}=\frac{x^2+x+1}{3}\Rightarrow p(x+1)=x^2+x+1$
Replace $x\to x-1$
$p\left(x\right)=(x-1{)}^2+(x-1)+1=x^2-x+1$
$I={\int }_0^1{\tan }^{-1}(x^2-x+1)\cdot {\tan }^{-1}\sqrt{\frac{x}{1-x}}dx$
$I={\int }_0^1{\tan }^{-1}(x^2-x+1)\cdot {\tan }^{-1}\sqrt{\frac{1-x}{x}}dx(\because {\int }_a^{b}f(x)dx={\int }_a^{b}f(a+b-x\left)dx\right)$
$2I={\int }_0^1{\tan }^{-1}(x^2-x+1)\cdot \frac{\pi }{2}dx2I={\int }_0^1\frac{\pi }{2}(\frac{\pi }{2}-ta{n}^{-1}(\frac{1}{x^2-x+1}\left)\right)dx$
$=\frac{{\pi }^2}{4}-\frac{\pi }{2}[{\int }_0^1(ta{n}^{-1}x+ta{n}^{-1}(1-x)dx]$
$=\frac{{\pi }^2}{4}-\frac{\pi }{2}\times 2{\int }_0^{1}ta{n}^{-1}xdx$
$2I=\frac{{\pi }^2}{4}-\pi (\frac{\pi }{4}-\frac{1}{2}\cdot ln2)I=\frac{\pi }{4}ln2$