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Question
Evaluate the following integrals : `int (3x + 4)/(x^2 + 6x + 5).dx`
Solution
Let I = `int (3x + 4)/(x^2 + 6x + 5).dx`
Let 3x + 4 = `"A"[d/dx(x^2 + 6x + 5)] + "B"`
= A(2x + 6) + B
∴ 3x + 4 = 2Ax + (6A + B)
Comparing the coefficient of x and constant on both sides, we get
2A = 3 and 6A + B = 4
∴ `"A" = (3)/(2) and 6(3/2) + "B"` = 4
∴ B = – 5
∴ 3x + 4 = `(3)/(2)(2x + 6) - 5`
∴ I = `int (3/2(2x + 6) - 5)/(x^2 + 6x + 5).dx`
= `(3)/(2) int (2x + 6)/(x^2 + 6x + 5).dx - 5 int (1)/(x^2 + 6x + 5).dx`
= `(3)/(2)"I"_1 - 5"I"_2`
I1 is of the type `int (f'(x))/f(x).dx = log|f(x)| + c`
∴ `"I"_1 = log|x^2 + 6x + 5| + c_1`
I2 = `int (1)/(x^2 + 6x + 5).dx`
= `int (1)/((x^2 + 6x + 9) - 4).dx`
= `int (1)/((x + 3)^2 - 2^2).dx`
= `(1)/(2 xx 2)log|(x + 3 - 2)/(x + 3 + 2)| + c_2`
= `(1)/(4)log|(x + 1)/(x + 5)| + c_2`
∴ I = `(3)/(2)log|x^2 + 6x+ 5| - (5)/(4)log|(x + 1)/(x + 5)| + c`, where c = c + c2.
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