Question

$2\int \frac{t^2+1}{t^4+t^2+1.dt}$

• A.
• B.
• C.
• D.

Answer 1

The correct option is A

We can see that 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
So, it'll be $=2.\frac{1}{\sqrt{3}}ta{n}^{-1}\left(\frac{u}{\sqrt{3}}\right)+C$
On substituting u = $t-\frac{1}{t}$, the integral becomes
$2.\frac{1}{\sqrt{3}}ta{n}^{-1}\left(\frac{t-\frac{1}{t}}{\sqrt{3}}\right)+C$
=$\frac{2}{\sqrt{3}}ta{n}^{-1}\left(\frac{t^2-1}{\sqrt{3}t}\right)+C$