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
⇒ $∫\frac {4x+1}{\sqrt {2x^2+x-3}}\ dx$ = $∫\frac {1}{\sqrt t} dt$
= $2\sqrt t+C$
= $2\sqrt 2x^2+x-3+C$