Evaluate the integral:
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