Question

Evaluate:

`int (3"e"^(2x) - 2"e"^x)/("e"^(2x) + 2"e"^x - 8)"d"x`

Answer 1

Let I = `int (3"e"^(2x) - 2"e"^x)/("e"^(2x) + 2"e"^x - 8)"d"x`

I = `int ((3"e"^x - 2)"e"^x "d"x)/("e"^(2x) + 2"e"^x - 8)`

I = `int (3"t" - 2)/("t"^2 + 2"t" - 8)"dt"`

Put e^x = t

e^x dx = dt   ...(1)

I = `int (3"t" - 2)/(("t" + 4)("t" - 2))"dt"`   ...(1)

Now `(3"t" - 2)/(("t" + 4)("t" - 2)) = "A"/("t" + 4) + "B"/("t" - 2)`  ...(2)

∴ A = `((3"t" - 2)/("t" - 2))_("t"  =  - 4)`

= `(3(-4) - 2)/(-4 - 2)`

= `14/6`

= `7/3`

B = `((3"t" - 2)/("t" + 4))_("t"  =  2)`

= `(3 xx 2 - 2)/(2 + 4)`

= `4/6`

= `2/3`

By (2) `(3"t" - 2)/(("t" + 4)("t" - 2)) = 7/(3("t" + 4)) + 2/(3("t" - 2))`

`\implies int (3"t" - 2)/(("t" + 4)("t" - 2))"dt" = 7/3 int 1/("t" + 4)"dt" + 2/3 int 1/("t" - 2)"dt"`

`\implies` I = `7/3log("t" + 4) + 2/3log("t" - 2) + "C"`

`\implies` I = `7/3log("e"^x + 4) + 2/3log("e"^x - 2) + "C"`.