Prove

Question:

$\frac{x+2}{\sqrt{x^{2}+2 x+3}}$

Solution:

$\int \frac{(x+2)}{\sqrt{x^{2}+2 x+3}} d x=\frac{1}{2} \int \frac{2(x+2)}{\sqrt{x^{2}+2 x+3}} d x$

$=\frac{1}{2} \int \frac{2 x+4}{\sqrt{x^{2}+2 x+3}} d x$

$=\frac{1}{2} \int \frac{2 x+2}{\sqrt{x^{2}+2 x+3}} d x+\frac{1}{2} \int \frac{2}{\sqrt{x^{2}+2 x+3}} d x$

$=\frac{1}{2} \int \frac{2 x+2}{\sqrt{x^{2}+2 x+3}} d x+\int \frac{1}{\sqrt{x^{2}+2 x+3}} d x$

Let $I_{1}=\int \frac{2 x+2}{\sqrt{x^{2}+2 x+3}} d x$ and $I_{2}=\int \frac{1}{\sqrt{x^{2}+2 x+3}} d x$

$\therefore \int \frac{x+2}{\sqrt{x^{2}+2 x+3}} d x=\frac{1}{2} I_{1}+I_{2}$   ...(1)

Then, $I_{1}=\int \frac{2 x+2}{\sqrt{x^{2}+2 x+3}} d x$

Let $x^{2}+2 x+3=t$

$\Rightarrow(2 x+2) d x=d t$

$I_{1}=\int \frac{d t}{\sqrt{t}}=2 \sqrt{t}=2 \sqrt{x^{2}+2 x+3}$   ...(2)

$I_{2}=\int \frac{1}{\sqrt{x^{2}+2 x+3}} d x$

$\Rightarrow x^{2}+2 x+3=x^{2}+2 x+1+2=(x+1)^{2}+(\sqrt{2})^{2}$

$\therefore I_{2}=\int \frac{1}{\sqrt{(x+1)^{2}+(\sqrt{2})^{2}}} d x=\log (x+1)+\sqrt{x^{2}+2 x+3} \mid$   ...(3)

Using equations (2) and (3) in (1), we obtain

$\int \frac{x+2}{\sqrt{x^{2}+2 x+3}} d x=\frac{1}{2}\left[2 \sqrt{x^{2}+2 x+3}\right]+\log \left|(x+1)+\sqrt{x^{2}+2 x+3}\right|+\mathrm{C}$

$=\sqrt{x^{2}+2 x+3}+\log \left|(x+1)+\sqrt{x^{2}+2 x+3}\right|+\mathrm{C}$