Question

Evaluate
$\int \frac{x+1}{x^2+3x+12}dx$

Answer 1

$\int \frac{x+1}{x^2+3x+12}dx$

$=\frac{1}{2}\int \frac{2(x+1)}{x^2+3x+12}dx$

$=\frac{1}{2}\int \frac{2x+2+1-1}{x^2+3x+12}dx$

$\Rightarrow \int \frac{x+1}{x^2+3x+12}dx=\frac{1}{2}[\underset{\left(u\right)}{\int \frac{2x+3}{x^2+3x+12}dx}-\underset{\left(v\right)}{\int \frac{1}{x^2+3x+12}}].....\left(1\right)$

Solving for $u$ and $v$, we have

$u=\int \frac{2x+3}{x^2+3x+12}dx$

Let $x^2+3+12=t\Rightarrow (2x+3)dx=dt$

$\therefore u=\int \frac{dt}{t}=\log t+C$

Substituting the value of $t$, we get

$u=\log (x^2+3x+12)+C_1$

Now,

$v=\int \frac{1}{x^2+3x+12}dx$

$\Rightarrow v=\int \frac{1}{{(x+\frac{3}{2})}^2-\frac{9}{4}+12}dx=\int \frac{1}{{(x+\frac{3}{2})}^2+{\left(\frac{\sqrt{39}}{2}\right)}^2}dx$

$\therefore v=\frac{2}{\sqrt{39}}{\tan }^{-1}\left(\frac{x+\frac{3}{2}}{\frac{\sqrt{39}}{2}}\right)$

$\Rightarrow v=\frac{2}{\sqrt{39}}{\tan }^{-1}\left(\frac{2x+3}{\sqrt{39}}\right)+C_2$

Substituting the value of $x$ in equation $\left(1\right)$, we have

$\Rightarrow \int \frac{x+1}{x^2+3x+12}dx$ $=\frac{1}{2}[\log (x^2+3x+12)-\frac{2}{\sqrt{39}}{\tan }^{-1}\left(\frac{2x+3}{\sqrt{39}}\right)+C_1-c_2]=\frac{1}{2}\log (x^2+3x+12)-\frac{1}{\sqrt{39}}{\tan }^{-1}\left(\frac{2x+3}{\sqrt{39}}\right)+C$