Question

If $I=\int \frac{2x+3}{(x^2+2x+3)\sqrt{x^2+2x+4}}dx$, then I equals

• A. $\log \left|\frac{\sqrt{x^2+2x+4}-1}{\sqrt{x^2+2x+4}+1}\right|+C$
• B. $\log \left|\frac{\sqrt{x^2+2x+4}-1}{\sqrt{x^2+2x+4}+1}\right|+{\tan }^{-1}\left(\frac{x+2}{3}\right)+C$
• C. $\log {\tan }^{-1}\left(\frac{\sqrt{x+3}}{2}\right)+C$
• D. none of these

Answer 1

The correct option is D none of these

$I=\int \frac{2x+3}{(x^2+2x+3)\sqrt{x^2+2x+4}}dx$
Let $I=2I_1+I_2$
Where
$I_1=\int \frac{(x+1)dx}{({(x+1)}^2+2)\sqrt{{(x+1)}^2+3}}$
Substitute ${(x+1)}^2+3=t$
$I_1=\int \frac{tdt}{(t^2-1)t}=\frac{1}{2}\log \left|\frac{t-1}{t+1}\right|=\frac{1}{2}\log \left|\frac{\sqrt{x^2+2x+4}-1}{\sqrt{x^2+2x+4}+1}\right|+c$
And
$I_2=\int \frac{dx}{({(x+1)}^2+2)\sqrt{{(x+1)}^2+3}}$
Substitute $x+1=\frac{1}{t}$
$I_2=\int \frac{(-\frac{1}{t^2})t^{3}dt}{(1+2t^2)\sqrt{1+3t^2}}$
Substitute $1+3t=u^2$
$I_2=-\int \frac{udu}{(2u^2+1)u}=-\frac{1}{\sqrt{2}}{\tan }^{-1}\left(\frac{\sqrt{2(x^2+2x+4)}}{x+1}\right)$