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

Question

Evaluate the integral :

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

Answer 1

I = \(\int\frac{x-3}{x^2+2x-4}\)dx

As we can see that there is a term of x in numerator and derivative of x² is also 2x. 

So there is a chance that we can make substitution for x² + 2x –4 and I can be reduced to a fundamental integration.

As,

\(\frac{d}{dx}\)(x² + 2x - 4) = 2x + 1

∴ Let, 

x - 3 = A(2x + 1) + B 

⇒ x - 3 = 2Ax + A + 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 |x² + 2x - 4| + C ….eqn 2

As,

I₂ = \(\int\frac{1}{x^2+2x-4}\)dx

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 :