Question

If $\int \frac{dx}{(x+2)(x^2+1)}=aln(1+x^2)+b{\tan }^{-1}x+\frac{1}{5}ln|x+2|+C$ then

• A. $a=-\frac{1}{10},b=-\frac{2}{5}$
• B. $a=\frac{1}{10},b=-\frac{2}{5}$
• C. $a=-\frac{1}{10},b=\frac{2}{5}$
• D. $a=\frac{1}{10},b=\frac{2}{5}$

Answer 1

The correct option is C $a=-\frac{1}{10},b=\frac{2}{5}$

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

When$1=A(4+1)+0(Bx+C)A=\frac{1}{65}1=\frac{x^2+1}{5}+(Bx+C)(x+2)$

Let $x=01=\frac{1}{5}+C\left(2\right),C=\frac{2}{5}1=\frac{x^2+1}{5}+(Bx+\frac{2}{5})(x+2)$

Let $x=21=\frac{4+1}{5}+(2B+\frac{2}{5})(2+2)V=1+(2B+\frac{2}{5})4$

is $B=-\frac{1}{5}$

$=\int \frac{\frac{1}{5}}{x+2}+\int \frac{-\frac{1}{5}x+\frac{2}{5}}{x^2+1}dx=\frac{1}{5}\int \frac{dx}{x+2}-\frac{1}{5}\int \frac{xdx}{x^2+1}+\frac{2}{5}\int \frac{dx}{x^2+1}=\frac{1}{5}\ln |x+2|-\frac{1}{5}\times \frac{1}{2}\int \frac{2xdx}{x^2+1}+\frac{2}{5}{\tan }^{-1}x=\frac{1}{5}\ln |x+2|-\frac{1}{10}\ln |x^2+1|+\frac{2}{5}{\tan }^{-1}x+C$

So $a=-\frac{1}{10}$ &$b=\frac{2}{5}$

Answer $C$