Question

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

• A.
• B.
• C.
• D.

Answer 1

The correct option is C

This is an another form of special irrational algebraic functions. To solve this we’ll substitute $x+1=t^2$ or $x=t^2-1$
& dx = 2t.dt
$\int \frac{t^2-1+2}{\left(\right(t^2-1{)}^2+3(t^2-1)+3)\sqrt{t^2-1+1}}2t.dt$
$\int \frac{(t^2+1).2t}{\left(\right(t^2-1{)}^2+3(t^2-1)+3).t}$
$2\int \frac{t^2+1}{t^4+t^2+1.dt}$​
Now we are familiar with the form we have, this is $\int \frac{x^2\pm 1}{x^4+K{x}^2+1}dx$ form and to solve this we’ll have to divide numerator and denominator by $t^2$.
So, we'll have -
$2\int \frac{1+\frac{1}{t^2}}{t^2+1+\frac{1}{t^2}}dt$
$=2\int \frac{1+\frac{1}{t^2}}{(t-\frac{1}{t}{)}^2+3}dt$
Substitute $t-\frac{1}{t}=u$
$1+\frac{1}{t^2}$ .dt = du
So, the integral will be =
$2\int \frac{1}{(u{)}^2+(\sqrt{3}{)}^2}du$
Which is a standard form. And we can apply the corresponding formula

On substituting u = $t-\frac{1}{t}$ and using standard formula, the integral becomes
$2.\frac{1}{\sqrt{3}}ta{n}^{-1}\left(\frac{t-\frac{1}{t}}{\sqrt{3}}\right)+C$

On substituting $t=\sqrt{x+1}$
$\frac{2}{\sqrt{3}}ta{n}^{-1}\left(\frac{x}{\sqrt{3(x+1)}}\right)+C$