How do you integrate $int (e^x+e^-x)/(e^x-e^-x)dx$ ?

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Answer 1 · 2017-02-20T02:39:45

$int (e^x + e^-x)/(e^x - e^-x)dx = ln|sinhx| + C = ln|1/2(e^x -e^-x)| + C$

We know that

$•coshx = (e^x + e^-x)/2$
$•sinhx = (e^x- e^-x)/2$

This integral can be rewritten as

$int coshx/sinhx dx$ , where $coshx$ and $sinhx$ represent the hyperbolic trigonometric functions

Now use a substitution to solve. Let $u = sinhx$ . Just like with regular trigonometric functions, $du = coshx dx$ and $dx= (du)/coshx$ .

$int coshx/u * (du)/coshx$

$int 1/u du$

$ln|u| + C$

$ln|sinhx| + C$

If you wish, the answer can be written as

$ln|1/2(e^x - e^-x)| + C$

Hopefully this helps!

How do you integrate int (e^x+e^-x)/(e^x-e^-x)dxint (e^x+e^-x)/(e^x-e^-x)dx?