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

Answers (1)

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

By partial fractions

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

$x^2+x+1=A(x^2+1)+(Bx+C)(x+2)$

$=Ax^2+A+Bx^2+2Bx+Cx+2C$

$x^2+x+1=x^2(A+B)+x(2B+C)+A+2C\: \: -(i)$

On comparing coefficients of equation (i), we get

$1=A+B\: \: and\: \: 1=2B+C$

On putting $x=-2\: \: in\: (i)$ we get

$(-2)^2+(-2)+1=(-2)^2A+A$

$3=4A+A$

$\frac{3}{5}=A\: \: Then\: \: 1=\frac{3}{5}+B$

$B=\frac{2}{5}\: \: \: and\: \: C=\frac{1}{5}$

Hence,

$\frac{(x^2+x+1)}{(x+2)(x^2+1)}=\frac{3}{5(x+2)}+\frac{\frac{2}{5}x+\frac{1}{5}}{x^2+1}$

Then, $\Rightarrow \int \frac{(x^2+x+1)}{(x+2)(x^2+1)}dx=\frac{3}{5}\int \frac{1}{(x+2)}dx+\frac{2}{5}\int \frac{x}{x^2+1}dx+\frac{1}{5}\int \frac{1}{x^2+1}dx$

$\Rightarrow \frac{3}{5}\log(x+2)+\frac{1}{5}\int \frac{2xdx}{x^2+1}+\frac{1}{5}\int \frac{\mathrm{d} x}{\mathrm{} x^2+1}$

$\Rightarrow \frac{3}{5}\log(x+2)+\frac{1}{5}\log \left | x_2+1 \right | +\frac{1}{5}\tan^{-1}x+c$

Posted by

Ravindra Pindel

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