Question

Integration of 3x^2/x^6+1​

Answer 1

Answer:

$\displaystyle \longrightarrow \tan^{-1}(x^3)+C$

Step-by-step explanation:

Let :

$\displaystyle \text{I} = \int\limits {\frac{3x^2}{x^6+1} } \, dx$

Using substitution method :

$\displaystyle u = x^3$

Diff. w.r.t. x we get :

$\displaystyle \longrightarrow \frac{du}{dx} =3x^2$

$\displaystyle \longrightarrow dx=\left(\frac{1}{3x^2}\right).du$

Now :

$\displaystyle \text{I} = \int\limits {\frac{3x^2}{((x^3)^2+1} } \, dx$

Putting value here we get :

$\displaystyle \text{I} = \int\limits {\frac{1}{u^2+1} } \, du$

We know :

$\displaystyle \int\limits {\frac{1}{x^2+1} } \, dx = \tan^{-1}(x)$

$\displaystyle \longrightarrow \text{I} = \tan^{-1}(u)+C$

$\displaystyle \sf But \ u=x^3$

$\displaystyle \longrightarrow \text{I} = \tan^{-1}(x^3)+C$

Hence we get required answer!