Integrate the function: \(\frac{x+2}{\sqrt {x^2+2x+3}}\)
Solution and Explanation
∫\(\frac{x+2}{\sqrt{x^2+2x+3}}\)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
\(∫\frac{x+2}{\sqrt{x^2+2x+3}} dx = \frac{1}{2}[2\sqrt{x^2+2x+3}]+log|(x+1)+\sqrt{x^2+2x+3}|+C\)
= \(\sqrt{x^2+2x+3}+log|(x+1)+\sqrt{x^2+2x+3}|+C\)
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Concepts Used:
Integrals of Some Particular Functions
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.
Integrals of Some Particular Functions:
- ∫1/(x2 – a2) dx = (1/2a) log|(x – a)/(x + a)| + C
- ∫1/(a2 – x2) dx = (1/2a) log|(a + x)/(a – x)| + C
- ∫1/(x2 + a2) dx = (1/a) tan-1(x/a) + C
- ∫1/√(x2 – a2) dx = log|x + √(x2 – a2)| + C
- ∫1/√(a2 – x2) dx = sin-1(x/a) + C
- ∫1/√(x2 + a2) dx = log|x + √(x2 + a2)| + C
These are tabulated below along with the meaning of each part.
