Question

Integrate the function: $\frac{x+2}{\sqrt {x^2+2x+3}}$

Answer 1

∫$\frac{x+2}{\sqrt{x^2+2x+3}}$dx = $\frac{1}{2}$ ∫$\frac{2(x+2)}{\sqrt{x^2+2x+3}}$ dx

=$\frac{1}{2}$ ∫$\frac{2x+4}{\sqrt{x^2+2x+3}}$ dx

=$\frac{1}{2}$ ∫$\frac{2x+2}{\sqrt{x^2+2x+3}}$dx +$\frac{1}{2}$ ∫$\frac{2}{\sqrt{x^2+2x+3}}$ dx

=$\frac{1}{2}$∫$\frac{2x+2}{\sqrt{x^2+2x+3}}$ dx + ∫$\frac{1}{\sqrt{x^2+2x+3}}$ dx

Let I1 = ∫$\frac{2x+2}{\sqrt{x^2+2x+3}}$ dx and I2 = ∫$\frac{1}{\sqrt{x^2+2x+3}}$ dx

∴ ∫$\frac{x+2}{\sqrt{x^2+2x+3}}$dx = $\frac{1}{2} I_1+I_2$ ...(1)

Then, I1 = ∫$\frac{2x+2}{\sqrt{x^2+2x+3}}$ dx

Let x2 + 2x +3 = t

⇒ (2x + 2) dx =dt

I1 = ∫$\frac{dt}{√t} = 2√t = 2\sqrt{x^2+2x+3}$ ...(2)

I2 = ∫$\frac{1}{\sqrt{x^2+2x+3}} dx$

⇒ $x^2+2x+3 = x^2+2x+1+2 = (x+1)^2 + (√2)^2$

$∴ I_2 = ∫\frac{1}{√(x+1)^2+(√2)^2}dx = log|(x-1)+\sqrt{x^2+2x+3 } ...(3)$

Using equations (2) and (3) in (1), we obtain

$∫\frac{x+2}{\sqrt{x^2+2x+3}} dx = \frac{1}{2}[2\sqrt{x^2+2x+3}]+log|(x+1)+\sqrt{x^2+2x+3}|+C$

= $\sqrt{x^2+2x+3}+log|(x+1)+\sqrt{x^2+2x+3}|+C$