Question

Evaluate: $\int \frac{x+2}{\sqrt{x^2+5x+6}}dx.$

Answer 1

Let $I=\int \frac{x+2}{\sqrt{x^2+5x+6}}dx$

Put $x+2=\lambda \left(\frac{d}{dx}\right(x^2+5x+6\left)\right)+\mu$

$x+2=2\lambda x+5\lambda +\mu$

Comparing coefficients of x both sides

$1=2\lambda \Rightarrow \lambda =\frac{1}{2}$

Comparing constant terms both sides,

$2=5\lambda +\mu$

or, $2=5\left(\frac{1}{2}\right)+\mu$

$\mu =2-\frac{5}{2}=\frac{-1}{2}$

Therefore, $\int \frac{x+2}{\sqrt{x^2+5x+6}}dx=\int \frac{\frac{1}{2}(2x+5)-\frac{1}{2}}{\sqrt{x^2+5x+6}}dx$ $(x+2=\lambda (2x+5)+\mu )$

$I=\int \frac{\frac{1}{2}(2x+5)}{\sqrt{x^2+5x+6}}dx-\frac{1}{2}\int \frac{dx}{\sqrt{x^2+5x+6}}$

Therefore, $I=I_1-I_2$---- (1)

$I_1=\int \frac{\frac{1}{2}(2x+5)}{\sqrt{x^2+5x+6}}dx$

Put $x^2+5x+6=t$

$\therefore (2x+5)dx=dt$

$=\frac{1}{2}\int \frac{dt}{\sqrt{t}}=\frac{1}{2}\left(\frac{t^{\frac{-1}{2}+1}}{\frac{-1}{2}+1}\right)+C$

$=t^{\frac{1}{2}}+C=\sqrt{t}+C\Rightarrow \sqrt{x^2+5x+6}+C$

$I_2=\frac{1}{2}\int \frac{dx}{\sqrt{x^2+5x+6}}$

$\frac{1}{2}\int \frac{dx}{\sqrt{x^2+5x+\frac{25}{4}-\frac{25}{4}+6}}=\frac{1}{2}\int \frac{dx}{\sqrt{{(x+\frac{5}{2})}^2-{\left(\frac{1}{2}\right)}^2}}$

$\frac{1}{2}.log[(x+\frac{5}{2})+\sqrt{{(x+\frac{5}{2})}^2-{\left(\frac{1}{2}\right)}^2}]+C$

$\frac{1}{2}.log[(x+\frac{5}{2})+\sqrt{x^2+5x+6}]+C$

Substituting the values, of $I_1$ and $I_2$ in (1),

$I=\sqrt{x^2+5x+6}+\frac{1}{2}.log[(x+\frac{5}{2})+\sqrt{x^2+5x+6}]+C$