Integrate the function: \(\frac {4x+1}{\sqrt {2x^2+x-3}}\)
Solution and Explanation
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
⇒ \(∫\frac {4x+1}{\sqrt {2x^2+x-3}}\ dx\) = \(∫\frac {1}{\sqrt t} dt\)
= \(2\sqrt t+C\)
= \(2\sqrt 2x^2+x-3+C\)
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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.
