Evaluate the integral : ∫ x^3/(x^4+x^2+1) dx ​

Question

Evaluate the integral :

\(\int\frac{x^3}{x^4+x^2+1}\) dx

Answer 1

Let, I = \(\int\frac{x^3}{x^4+x^2+1}\) dx

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

If we assume x² to be an another variable, we can simplify the integral as derivative of x² i.e. x is present in numerator. 

Let, 

x² = u 

⇒ 2x dx = du 

⇒ x dx = 1/2 du

∴ I = \(\frac{1}{2}\int\frac{u}{u^2+u+1}\)du

As,

\(\frac{d}{du}\)(u² + u +1) = 2u +1

∴ Let, 

u = A(2u + 1) + B 

⇒ u = 2Au + A + B 

On comparing both sides – 

We have, 

2A = 1 

⇒ A = 1/2 A + B = 0 

⇒ B = –A = –1/2 

Hence,

Now,

I = I₁ + I₂ ….eqn 1 

We will solve I₁ and I₂ individually. 

As, 

I₁ = \(\frac{1}{4}\int\frac{(2u+1)}{u^2+u+1}\) du

Let v = u² + u + 1 

⇒ dv = (2u + 1)du 

∴ I₁ reduces to \(\frac{1}{4}\int\frac{dv}{v}\)

Hence,

As, 

I₂ = \(-\frac{1}{4}\int\frac{1}{u^2+u+1}\)du and 

we don’t have any derivative of function present in denominator. 

∴ we will use some special integrals to solve the problem. 

As denominator doesn’t have any square root term. 

So one of the following two integrals will solve the problem.

Now, 

We have to reduce I₂ such that it matches with any of above two forms. 

We will make to create a complete square so that no individual term of x is seen in denominator.