Question

Integrate the following w.r.t. x : `(5*e^x)/((e^x + 1)(e^(2x) + 9)`

Answer 1

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

Put e^x = t
∴ e^x.dx = dt

∴ I = `5 int (1)/((t + 1)(t^2 + 9))*dt`

Let `(1)/((t + 1)(t^2 + 9)) = "A"/(t + 1) + "Bt + C"/(t^2 + 9)`
∴ 1 = A(t² + 9) + (Bt + C)(t + 1)
Put t + 1 = 0, i.e. t = – 1, we get

1 = A(1 + 9) + C(0)

∴ A = `(1)/(10)`
Put t = 0, we get
1 = A(9) + C(1)

∴ C = 1 – 9A = `1 - (9)/(10) = (1)/(10)`
Comparing coefficients of t2 on both the sides, we get
0 = A + B
∴ B = – A = `-(1)/(10)`

∴ `(1)/((t + 1)(t^2 + 9)) = ((1/10))/(t + 1) + ((-1/10t + 1/10))/(t^2 + 9)`

∴ I = `5 int [((1/10))/(t + 1) + ((-1/10t + 1/10))/(t^2 + 9)]*dt`

= `(1)/(2) int (1)/(t + 1)*dt - (1)/(2) int t/(t^2 + 9)*dt + (1)/(2) int t/(t^2 + 9)*dt`

= `(1)/(2)log|t + 1| - (1)/(4) int (2t)/(t^2 + 9)*dt + (1)/(2).(1).(3)tan^-1(t/3)`

= `(1)/(2)log|t + 1| - (1)/(4) int (d/dt(t^2 + 9))/(t^2 + 9)*dt + (1)/(6)tan^-1(t/3)`

= `(1)/(2)log|t + 1| - (1)/(4)log|t^2 + 9| + (1)/(6)tan^-1(t/3) + c`

= `(1)/(2)log|e^x + 1| - (1)/(4)log|e^(2x) + 9| + (1)/(6)tan^-1(e^x/3) + c`.