Question

The value of the integral $\int \frac{x^2+x+1}{(x+2)(x^2+1)}dx$
(where $m$ is integration constant)

• A. $\frac{2}{5}\ln |x+2|+\frac{1}{5}\ln |x^2+1|+\frac{1}{5}{\tan }^{-1}x+m$
• B. $\frac{3}{5}\ln |x+2|+\frac{1}{5}\ln |x^2+1|+\frac{1}{5}{\tan }^{-1}x+m$
• C. $\frac{1}{5}\ln |x+2|+\frac{1}{5}\ln |x^2+1|+\frac{1}{5}{\tan }^{-1}x+m$
• D. $\frac{3}{5}\ln |x+2|-\frac{1}{5}\ln |x^2+1|+\frac{1}{5}{\tan }^{-1}x+m$

Answer 1

The correct option is B $\frac{3}{5}\ln |x+2|+\frac{1}{5}\ln |x^2+1|+\frac{1}{5}{\tan }^{-1}x+m$

$\frac{x^2+x+1}{(x^2+1)(x+2)}=\frac{A}{x+2}+\frac{Bx+C}{(x^2+1)}$
Therefore,
$x^2+x+1=A(x^2+1)+(Bx+C)(x+2)$
Equating the coefficients of $x^2,x$ and of constant term of both sides, we get
$A+B=12B+C=1A+2C=1$
Solving these equations, we get
$A=\frac{3}{5},B=\frac{2}{5},C=\frac{1}{5}$

Now,
$\frac{x^2+x+1}{(x^2+1)(x+2)}=\frac{3}{5(x+2)}+\frac{1}{5}\left(\frac{2x+1}{x^2+1}\right)$
Therefore,
$\int \frac{x^2+x+1}{(x^2+1)(x+2)}dx=\frac{3}{5}\int \frac{dx}{x+2}+\frac{1}{5}\int \frac{2x}{x^2+1}dx+\frac{1}{5}\int \frac{1}{x^2+1}dx$
$=\frac{3}{5}\ln |x+2|+\frac{1}{5}\ln |x^2+1|+\frac{1}{5}{\tan }^{-1}x+m$