I =

As we can see that there is a term of x³ in numerator and derivative of x⁴ is also 4x³. So there is a chance that we can make substitution for x⁴ + c² and I can be reduced to a fundamental integration but there is also a x term present. So it is better to break this integration.

I = = I₁ + I₂ …eqn 1

I₁ =

As,

To make the substitution, I₁ can be rewritten as

∴ Let, x⁴ + c² = u

⇒ du = 4x³ dx

I₁ is reduced to simple integration after substituting u and du as:

∴ I₁ = …eqn 2

As,

I₂ =

∵ we have derivative of x² in numerator and term of x² in denominator. So we can apply method of substitution here also.

As, I₂ =

Let, x² = v

⇒ dv = 2x dx

∵ I₂ = =

As denominator doesn’t have any square root term. So one of the following two integrals will solve the problem.

I₂ matches with

∴ I₂ =

⇒ I₂ = …eqn 3

From eqn 1, we have:

I = I₁ + I₂

Using eqn 2 and 3, we get –

I = …..ans