Question

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

Answer 1

Let $x+2=A\frac{d}{dx}(x^2+2x+3)+B\Rightarrow x+2=A(2x+2)+B$
$\Rightarrow x+2=2Ax+(2A+B)$
On equating the coefficient of x and constant term on both sides, we get
$2A=1\Rightarrow A=\frac{1}{2}and2A+B=2\Rightarrow 2\times \frac{1}{2}+B=2\Rightarrow B=2-1=1\Rightarrow x+2=\frac{1}{2}(2x+2)+1\therefore \int \frac{x+2}{\sqrt{x^2+2x+3}}dx=\int \frac{\frac{1}{2}(2x+2)+1}{\sqrt{x^2+2x+3}}$
Let $I_1=\int \frac{2x+2}{\sqrt{x^2+2x+3}}dxand{I}_2=\int \frac{dx}{\sqrt{x^2+2x+3}}$
Then $\int \frac{x+2}{\sqrt{x^2+2x+3}}dx=\frac{1}{2}{I}_1+I_2.....\left(i\right)$
Now, $I_1=\int \frac{2x+2}{\sqrt{x^2+2x+3}}dx$
Let $x^2+2x+3=t\Rightarrow (2x+2)dx=dt$
$\therefore I_1=\int \frac{dt}{\sqrt{t}}=\int t^{-\frac{1}{2}}dt=\frac{t^{(-\frac{1}{2})+1}}{-\frac{1}{2}+1}+C_1=2\sqrt{x^2+2x+3}+C_1$
and $I_2=\int \frac{dx}{\sqrt{x^2+2x+3}}=\int \frac{dx}{\sqrt{x^2+2x+(1{)}^2+3-(1{)}^2}}=\int \frac{dx}{\sqrt{(x+1{)}^2+(\sqrt{2}{)}^2}}$
Let $x+1=t\Rightarrow dx=dt$
$I_2=\int \frac{dt}{\sqrt{t^2+(\sqrt{2}{)}^2}}=log|t+\sqrt{t^2+2}|+C_2[\because \int \frac{dx}{\sqrt{x^2+a^2}}=log|x+\sqrt{x^2+a^2}|]=log|x+1+\sqrt{(x+1{)}^2+2}|+C_2=log|x+1+\sqrt{x^2+2x+3}|+C_2$
On putting the values of $I_1$ and $I_2$ in Eq. (i), we get
$\int \frac{x+2}{\sqrt{x^2+2x+3}}dx=\frac{1}{2}\left[2\sqrt{x^2+2x+3}\right]+log|x+1+\sqrt{x^2+2x+3}|+C[\therefore \frac{1}{2}{C}_1+C_2=C]=\sqrt{x^2+2x+3}+log|x+1+\sqrt{x^2=2x+3}|+C$