∫\(\frac{x+2}{\sqrt{x^2+2x+3}}\)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
\(∫\frac{x+2}{\sqrt{x^2+2x+3}} dx = \frac{1}{2}[2\sqrt{x^2+2x+3}]+log|(x+1)+\sqrt{x^2+2x+3}|+C\)
= \(\sqrt{x^2+2x+3}+log|(x+1)+\sqrt{x^2+2x+3}|+C\)
What is the Planning Process?
There are many important integration formulas which are applied to integrate many other standard integrals. In this article, we will take a look at the integrals of these particular functions and see how they are used in several other standard integrals.
These are tabulated below along with the meaning of each part.