# Show that

Question:

$\frac{5 x+3}{\sqrt{x^{2}+4 x+10}}$

Solution:

Let $5 x+3=A \frac{d}{d x}\left(x^{2}+4 x+10\right)+B$

$\Rightarrow 5 x+3=A(2 x+4)+B$

Equating the coefficients of x and constant term, we obtain

$2 A=5 \Rightarrow A=\frac{5}{2}$

$4 A+B=3 \Rightarrow B=-7$

$\therefore 5 x+3=\frac{5}{2}(2 x+4)-7$

$\Rightarrow \int \frac{5 x+3}{\sqrt{x^{2}+4 x+10}} d x=\int \frac{\frac{5}{2}(2 x+4)-7}{\sqrt{x^{2}+4 x+10}} d x$

$=\frac{5}{2} \int \frac{2 x+4}{\sqrt{x^{2}+4 x+10}} d x-7 \int \frac{1}{\sqrt{x^{2}+4 x+10}} d x$

Let $I_{1}=\int \frac{2 x+4}{\sqrt{x^{2}+4 x+10}} d x$ and $I_{2}=\int \frac{1}{\sqrt{x^{2}+4 x+10}} d x$

$\therefore \int \frac{5 x+3}{\sqrt{x^{2}+4 x+10}} d x=\frac{5}{2} I_{1}-7 I_{2}$    ...(1)

Then, $I_{1}=\int \frac{2 x+4}{\sqrt{x^{2}+4 x+10}} d x$

Let $x^{2}+4 x+10=t$

$\therefore(2 x+4) d x=d t$

$\Rightarrow I_{1}=\int \frac{d t}{t}=2 \sqrt{t}=2 \sqrt{x^{2}+4 x+10}$    ...(2)

$I_{2}=\int \frac{1}{\sqrt{x^{2}+4 x+10}} d x$

$=\int \frac{1}{\sqrt{\left(x^{2}+4 x+4\right)+6}} d x$

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

$=\log \left|(x+2) \sqrt{x^{2}+4 x+10}\right|$    ...(3)

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

$\int \frac{5 x+3}{\sqrt{x^{2}+4 x+10}} d x=\frac{5}{2}\left[2 \sqrt{x^{2}+4 x+10}\right]-7 \log \left|(x+2)+\sqrt{x^{2}+4 x+10}\right|+C$

$=5 \sqrt{x^{2}+4 x+10}-7 \log (x+2)+\sqrt{x^{2}+4 x+10} \mid+C$