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प्रश्न
Evaluate:
`int (5e^x)/((e^x + 1)(e^(2x) + 9)) dx`
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उत्तर
Let I = `int (5e^x)/((e^x + 1)(e^(2x) + 9)) dx`
Put ex = t
⇒ exdx = dt
∴ I = `int 5/((t + 1)(t^2 + 9))dt`
Let `5/((t + 1)(t^2 + 9))`
= `A/(t + 1) + (Bt + C)/(t^2 + 9)`
∴ 5 = A(t2 + 9) + (Bt + C)(t + 1) ...(i)
Putting t = –1 in (i), we get
5 = A[(–1)2 + 9]
∴ 5 = 10A
∴ A = `1/2`
Putting t = 0 in (i), we get
5 = A(0 + 9) + (0 + C) (0 + 1)
∴ 5 = 9A + C
∴ 5 = `9(1/2) + "C"`
∴ C = `1/2`
Putting t = 1 in (i), we get
5 = A(12 + 9) + (B + C)(1 + 1)
∴ 5 = 10A + 2B + 2C
∴ 5 = `10(1/2) + 2B + 2(1/2)`
∴ – 1 = 2B
∴ B = `-1/2`
∴ `5/((t + 1)(t^2 + 9)) = (1/2)/(t + 1) + (1/2t + 1/2)/(t^2 + 9)`
∴ I = `int((1/2)/(t+ 1) + ((-1)/2t + 1/2)/(t^2 + 9)) dt`
= `1/2 [int 1/(t + 1) dt - int t/(t^2 + 9) dt + int 1/(t^2 + 9) dt]`
= `1/2[int 1/(t + 1) dt - 1/2 int (2t)/(t^2 + 9) dt + int 1/(t^2 + 3^2) dt]`
= `1/2 [log (t + 1) - 1/2 log (t^2 + 9) + 1/3tan^-1(t/3)] + c`
∴ I = `1/2 log (e^x + 1) - 1/4 log (e^(2x) + 9) + 1/6 tan^-1 ((e^x)/3) + c`
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