Question

${\int }_0^1\frac{2x^2+3x+3}{(x+1)(x^2+2x+2)}dx=$

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

Answer 1

The correct option is A $\frac{\pi }{4}+2ln2-{\tan }^{-1}2$

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

$\frac{2x^2+3x+3}{(x+1)(x^2+2x+2)}=\frac{A}{x+1}+\frac{Bx+C}{x^2+2x+2}$

$\frac{2x^2+3x+3}{(x+1)(x^2+2x+2)}=\frac{A(x^2+2x+2)+\left(Bx+C\right)\left(x+1\right)}{\left(x+1\right)\left(x^2+2x+2\right)}$

$2x^2+3x+3=A(x^2+2x+2)+\left(Bx+C\right)\left(x+1\right)$ ---- (i)

On putting $x=-1$ in (i), we get

$A=2$

and on putting $x=0$ in (i), we get

$C=-1$

Now put $x=1$ in (i), we get

$2+3+3=A(1+2+2)+2(B+C)$

put $A=2;C=-1$

$B=0$

Therefore,

${\int }_0^1\frac{2x^2+3x+3}{(x+1)(x^2+2x+2)}dx={\int }_0^1\frac{2}{x+1}dx-{\int }_0^1\frac{dx}{x^2+2x+2}$

$={\left[2\log \left(x+1\right)\right]}_0^1-{\int }_0^1\frac{dx}{x^2+2x+1+1}$

$={\left[2\log \left(x+1\right)\right]}_0^1-{\int }_0^1\frac{dx}{{\left(x+1\right)}^2+1}$

$={\left[2\log \left(x+1\right)-{\tan }^{-1}\left(x+1\right)\right]}_0^1$

$=\left(2\log 2-{\tan }^{-1}\left(2\right)\right)-\left(-\frac{\pi }{4}\right)$

$=\frac{\pi }{4}+2\log 2-{\tan }^{-1}\left(2\right)$