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Question
Evaluate the following integral:
`int ("d"x)/("e"^x + 6 + 5"e"^-x)`
Solution
`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 ex = t
Then exdx = 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"`
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