How do you find the indefinite integral of $int (e^(7x)) / (e^(14x) + 36) dx$ ?

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Answer 1 · 2015-09-27T03:23:39

Refer to explanation

Set $u=e^(7x)=>du=e^(7x)(7)dx=>du=7e^(7x)dx$

Hence the integral becomes

$int (e^(7x))/(e^(14x)+36)dx= 1/7 int (du)/(u^2+36)= 1/7*1/6int 1/((u/6)^2+1)du= 1/7*1/6*arctan(u/6)+c$

Because $u=e^7x$ we have that

$int (e^(7x))/(e^(14x)+36)dx=1/(7*6) arctan(e^(7x)/6)+c$

How do you find the indefinite integral of int (e^(7x)) / (e^(14x) + 36) dxint (e^(7x)) / (e^(14x) + 36) dx?