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