Question

Integrate the function: \(\frac {4x+1}{\sqrt {2x^2+x-3}}\)

Answer 1

Let 4x + 1 = A \(\frac {d}{dx}\)(2x²+x-3) + B
⇒ 4x+1 = A(4x+1) + B
⇒ 4x+1 = 4Ax + A + B
Equating the coefficients of x and constant term on both sides, we obtain 
4A = 4 ⇒ 1
A+B = 1 ⇒ B = 0
Let 2x² + x - 3 = t
∴ (4x+1) dx = dt
⇒ \(∫\frac {4x+1}{\sqrt {2x^2+x-3}}\  dx\) = \(∫\frac {1}{\sqrt t} dt\)
                                = \(2\sqrt t+C\)
                                = \(2\sqrt 2x^2+x-3+C\)