Question

Evaluate:

`int (5e^x)/((e^x + 1)(e^(2x) + 9)) dx`

Answer 1

Let I = `int (5e^x)/((e^x + 1)(e^(2x) + 9)) dx`

Put e^x = t

⇒ e^xdx = 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`