Question

`int (3"e"^(2"t") + 5)/(4"e"^(2"t") - 5)  "dt"`

Answer 1

Let I = `int (3"e"^(2"t") + 5)/(4"e"^(2"t") - 5)  "dt"`

Let 3e^2t + 5 = `"A"(4"e"^(2"t") - 5) + "B" "d"/"dt"(4"e"^(2"t") - 5)`

= 4Ae^2t – 5A + B(8e^2t)

∴ 3e^2t + 5 = (4A + 8B) e^2t – 5A

Comparing the coefficients of e^2t and constant term on both sides,

we get 4A + 8B = 3 and – 5A = 5

Solving these equations,

we get A = – 1 and B = `7/8`

∴ I = `int (-1(4"e"^(2"t") - 5) + 7/8(8"e"^(2"t")))/(4"e"^(2"t") - 5)  "dt"`

= `/int  "dt" + 7/8 int (8"e"^(2"t"))/(4"e"^(2"t") - 5)  "dt"`

∴ I = `-"t" + 7/8 log|4"e"^(2"t") - 5| + "c"`    ......`[because  int ("f'"(x))/("f"(x))  "d"x = log|"f"(x)| + "c"]`