Integrate the function: \(\frac{5x-2}{1+2x+3x^2}\)
Solution and Explanation
Let 5x-2 = A \(\frac{d}{dx}\)(1+2x+3x2)+B
⇒ 5x - 2 = A(2+6x)+B
Equating the coefficient of x and constant term on both sides, we obtain
5 = 6A ⇒ A\(\frac{5}{6}\)
2A+B = -2 ⇒ B= -11/3
∴ 5x-2 = 5/6(2+6x)+(-11/3)
⇒ ∫\(\frac{5x-2}{1+2x+3x^2}\)dx = ∫5/6(2+6x)-11/3/1+2x+3x2 dx
=5/6 ∫2+6x/1+2x+3x2 dx-11/3 ∫1/1+2x+3x2 dx
Let I1 = ∫2+6x/1+2x+3x2 dx and I2 = ∫1/1+2x+3x2 dx
∴ ∫\(\frac{5x-2}{1+2x+3x^2}\) dx = 5/6I1-11/3I2 ...(1)
I1 = ∫2+6x/1+2x+3x dx
Let 1+2x+3x2 = t
⇒ (2+6x)dx = dt
∴ I1 = ∫dt/t
I1 = log|t|
I1 = log|1+2x+3x2| ...(2)
I2 = ∫1/1+2x+3x2 dx
1+2x+3x2 can be written as 1+3(x2+2/3x).
Therefore,
1+3(x2+2/3x)
=1+3(x2+2/3x+1/9-1/9)
=1+3(x+1/3)2-1/3
=2/3+3(x+1/3)2
=3[(x+1/3)2+2/9]
=3[(x+1/3)2+(√2/3)2]
I2 = 1/3 ∫1/[(x+1/3)2+(√2/3)2]dx
=1/3[1/√2/3tan-1(3x+1)/√2)]
=1/√2tan-1(3x+1/√2) ...(3)
Substituting equations (2) and (3) in equation (1), we obtain
∫5x-2/1+2x+3x2 dx = 5/6[log|1+2x+3x2|]-11/3[1/√2tan-1(3x+1/√2)]+C
=5/6log|1+2x+3x2|-11/3√2tan-1(3x+1/√2)+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.
