Question

If $\underset{0}{\overset{1}{\int }}{\cot }^{-1}\left(1-x+x^2\right)dx=\lambda \underset{0}{\overset{1}{\int }}{\tan }^{-1}xdx$ then ' $\lambda$ ' is equal to

• A. 1
• B. 2
• C. 3
• D. 4

Answer 1

Answer: (B)

$\underset{0}{\overset{1}{\int }}{\cot }^{-1}\left(1-x+x^2\right)dx$
$=\underset{0}{\overset{1}{\int }}{\tan }^{-1}\left(\frac{1}{1-x+x^2}\right)dx=\underset{0}{\overset{1}{\int }}{\tan }^{-1}\left(\frac{x+(1-x)}{1-x(1-x)}\right)dx$
$=\underset{0}{\overset{1}{\int }}{\tan }^{-1}xdx+\underset{0}{\overset{1}{\int }}{\tan }^{-1}(1-x)dx$
$=\underset{0}{\overset{1}{\int }}{\tan }^{-1}xdx+\underset{0}{\overset{1}{\int }}{\tan }^{-1}[1-(1-x)]dx=2\underset{0}{\overset{1}{\int }}{\tan }^{-1}xdx$
$\Rrightarrow \lambda =2$