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प्रश्न
Integrate the function `(5x - 2)/(1 + 2x + 3x^2)`
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उत्तर
Let `I = int (5x - 2)/(1 + 2x + 3x^2) dx`
`5x - 2 = A d/dx (1 + 2x + 3x^2) + B`
= 5x - 2 = A (6x + 2) + B ....(i)
Comparing coefficients of x in (i), we get
6A = 5
`therefore A = 5/6`
-2 = 2A + B
`therefore B = -2 - 2A`
`= -2 -2 (5/6)`
`= -2 - 5/3`
`= - 11/3`
`therefore I = int (5/6 (2 + 6x) - 11/3)/(1 + 2x + 3x^2) dx`
`I = 5/6 int (2 + 6x)/(1 + 2x + 3x^2) dx`
`-11/3 int dx/(1 + 2x + 3x^2) dx`
` Let I = 5/6 I_1 - 11/3 I_2` .... (ii)
`therefore I_1 = 5/6 int (2 + 6x)/(1 + 2x + 3x^2) dx`
Put 1 + 2x + 3x2 = t
(2 + 6x) dx = dt
∴ `I_1 = int dt/t`
`= log |t| + C_1`
`= log |1 + 2x + 3x^2| + C_1` .....(iii)
`I_2 = int dx/(1 + 2x + 3x^2)`
`= 1/3 int dx/(x^2 + 2/3 x + 1/3)`
`= 1/3 int dx/((x^2 + 2/3 x + 1/9) + 1/3 - 1/9)`
`= 1/3 int dx/((x + 1/3)^2 + 2/9)`
`= 1/3 int dx/((x + 1/3)^2 + sqrt2/3)`
`= 1/3 . 1/(sqrt2/3) tan^-1 ((x + 1/3)/(sqrt2/3))`
`= 1/sqrt2 tan^-1 ((3x + 1)/sqrt2)` ... (iv)
From (ii), (iii) and (iv), we get
`I = 5/6 log (1 + 2x + 3x^2) - 11/3 xx 1/sqrt2 tan^-1 ((3x + 1)/sqrt2)`
`= 5/6 log (1 + 2x + 3x^2) - 11/(3 sqrt2) tan^-1 ((3x + 1)/sqrt2) + C`
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