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Question
Integrate the function:
`(5x)/((x+1)(x^2 +9))`
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Solution
Let I = `int (5x)/((x + 1)(x^2 + 9))`dx
`therefore (5x)/((x + 1)(x^2 + 9)) = A/(x + 1) + (Bx + C)/(x^2 + 9)`
`=> 5x = A(x^2 + 9) + (Bx + C)(x + 1)`
Putting x = -1 in equation (1),
- 5 = A(1 + 9)
⇒ - 5 = 10 A
`therefore A = - 5/10 = - 1/2`
From equation (1),
Comparing the coefficients of x2 and the constant term,
0 = A + B
⇒ B = - A = `1/2`
0 = 9A + C
⇒ C = - 9A = `9/2`
`therefore (5x)/((x + 1)(x^2 + 9)) = (- 1)/(2(x + 1)) + (1/2 x + 9/2)/(x^2 + 9)`
∴ `I = int(1/2)/(x + 1) dx + int (1/2 x + 9/2)/(x^2 + 9) dx`
`= -1/2 log (x + 1) + 1/4 int (2x)/(x^2 + 9) dx + 9/2 int dx/ (x^2 + 3^2) + C`
`= -1/2 log (x + 1) + 1/4 log (x^2 + 9) + 9/2 xx 1/3 tan^-1 x/3 + C`
`= -1/2 log (x + 1) + 1/4 log (x^2 + 9) + 3/2 tan^-1 x/3 + C`
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