Question

Integrate the following functions.
$\int \frac{5x+3}{x^2+4x+10}dx.$

Answer 1

Let $5x+3=A\frac{d}{dx}(x^2+4x+10)+B$
$\Rightarrow 5x+3=A(2x+4)+B\Rightarrow 5x+3=2Ax+4A+B$
On equating the coefficients of x and constant term on both sides, we get
$2A=5\Rightarrow A=\frac{5}{2}$
and $4A+B=3\Rightarrow B=-7\Rightarrow 5x+3=\frac{5}{2}(2x+4)-7$
$\therefore \int \frac{5x+3}{\sqrt{x^2+4x+10}}dx=\int \frac{\frac{5}{2}(2x+4)-7}{\sqrt{x^2+4x+10}}dx=\frac{5}{2}\int \frac{2x+4}{\sqrt{x^2+4x+10}}dx-7\frac{1}{\sqrt{x^2+4x+10}}dx$
Let $I_1=\int \frac{2x+4}{\sqrt{x^2+4x+10}}dxand{I}_2=\int \frac{1}{\sqrt{x^2+4x+10}}dx$
$\therefore \int \frac{5x+3}{\sqrt{x^2+4x+10}}dx=\frac{5}{2}{I}_1-7I_2..........\left(i\right)$
Now, $I_1=\int \frac{2x+4}{\sqrt{x^2+4x+10}}dx$
Let $x^2+4x+10=t\Rightarrow (2x+4)dx=dt\Rightarrow dx=\frac{dt}{2x+4}$
$\therefore I_1=\int \frac{2x+4}{\sqrt{t}}\times \frac{dt}{2x+4}=\int \frac{dt}{\sqrt{t}}=2\sqrt{t}+C_1=2\sqrt{x^2+4x+10}+C_1.......\left(ii\right)$
And, $I_2=\int \frac{1}{\sqrt{x^2+4x+10}}dx=\int \frac{1}{\sqrt{(x^2+4x+4)+6}}dx$
$=\int \frac{1}{\sqrt{(x+2{)}^2+(\sqrt{6}{)}^2}}dx=log|(x+2)+\sqrt{(x+2{)}^2+6}|+C_2.......\left(iii\right)[\because \int \frac{dx}{\sqrt{x^2+a^2}}=log|x+\sqrt{x^2+a^2}|]=log|x+2+\sqrt{x^2+4x+10}|+C_2$

On substituting the values of $I_1$ and $I_2$ from Eqs.(ii) and (iii)in Eq. (i), we get
$\int \frac{5x+3}{\sqrt{x^2+4x+10}}dx=\frac{5}{2}\left[2\sqrt{x^2+4x+10}\right]-7log|x+2+\sqrt{x^2+4x+10}|+C[\because \frac{5}{2}{C}_1-7C_2=C]=5\sqrt{x^2+4x+10}-7log|x+2+\sqrt{x^2+4x+10}|+C$