Question

Evaluate the following integral:

`int ("d"x)/("e"^x + 6 + 5"e"^-x)`

Answer 1

`int ("d"x)/("e"^x + 6 + 5"e"^-x) = int ("d"x)/("e"^x + 6 + 5/"e"^x)`

= `int ("e"^x  "d"x)/("e"^(2x) + 6"e"^x + 5)`

Let e^x = t

Then e^xdx = dt

So integral becomes = `ont "dt"/("t"^2 + 6"t" + 5)`

= `int "dt"/(("t" + 5)("t" + 1))`

Now = `1/(("t" + 5)("t" + 1))`

= `"A"/("t" + 5) + "B"/("t" + 1)`

1 = A(t + 1) + B(t + 5)

Put t = – 1

⇒ 4B = 1

B = `1/4`

Put t = – 5

⇒ – 4A = 1

A = ` 1/4`

So we have `1/4 int((-1)/("t" + 5) + 1/("t" + 1))  "dt"`

= `1/4 [- log("t" + 5) + log("t" + 1)] + "c"`

= `1/4 log |(("t" + 1))/(("t" + 5))| + "c"`

= `1/4 log |("e"^x + 1)/("e"^x + 5)| + "c"`