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

Question

Evaluate the integral:

\(\int\frac{x^3-3x}{x^4+2x^2-4}\)dx

Answer 1

Let, 

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

I =  \(\int\frac{x^2-3x}{(x^2)^2+2x^2-4}\)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 = \(\frac{1}{2}\int\frac{u-3}{u^2+2u-4}\)du

As,

\(\frac{d}{du}\)(u² + 2u - 4) = 2u + 2

∴ Let, 

u – 3 = A(2u + 2) + B 

⇒ u – 3 = 2Au + 2A + B 

On comparing both sides – 

We have, 

2A = 1 

⇒ A = 1/2 

2A + B = –3 

⇒ B = –3–2A = –4 

Hence,

On substituting value of u, we have: 

I₁ =  \(\frac{1}{2}\)log|u² + 2u - 4| + C….eqn 2

As, 

I₂ = \(\int\frac{1}{u^2+2u-4}\) 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.

From eqn 1 : 

I = I₁ – 4I₂ 

Using eqn 2 and eqn 3 :