Question

The value of ${\int }_0^1\frac{2x^2+3x+3}{(x+1)(x^2+2x+2)}dx$ is

• A. $\frac{\pi }{4}+2ln2-{\tan }^{-1}2$
• B. $\frac{\pi }{4}+2ln2-{\tan }^{-1}\frac{1}{3}$
• C. $2ln2-{\cot }^{-1}3$
• D. $-\frac{\pi }{4}+ln4+{\cot }^{-1}2$

Answer 1

Answer: (A), (C), (D)

$\frac{\pi }{4}+2ln2-{\tan }^{-1}2$
$2ln2-{\cot }^{-1}3$
$-\frac{\pi }{4}+ln4+{\cot }^{-1}2$

${\int }_0^1\frac{{2x}^2+3x+3}{(x+1)(x^2+2x+2)}dx={\int }_0^1(\frac{2}{x+1}-\frac{1}{x^2+2x+2})dx$
$=-{\int }_0^1\frac{1}{{(x+1)}^2}dx+2\int \frac{1}{x+1}dx={[-{\tan }^{-1}(x+1)+2\log (x+1)]}_0^1$
$=-{\tan }^{-1}2+2\log 2+{\tan }^{-1}1=2\log 2+\frac{\pi }{4}-{\tan }^{-1}2$
$=2\log 2-{\cot }^{-1}3=-\frac{\pi }{4}+\log 4+{\cot }^{-1}2$