How do you evaluate the integral of #[ (e^7x) / ((e^14x)+9) dx ]#?

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Answer 1 · 2015-05-04T04:20:03

#int{{ax}/{bx+c}}dx={ac}/b^2(b/cx-ln|b/cx+1|)+C#

where #a#, #b#, and #c# are constants. #C# is the constant of integration.
I got the above result by the following process:

#int {{ax}/{bx+c}}dx=int{a/{b}{b/cx}/{b/cx+1}}dx#

Let #u=b/cx#, Then #dx=c/bdu#.

By substituting, the integral in terms of #u# is,

#{ac}/b^2int{u/{u+1)}du={ac}/b^2int{1-1/{u+1}}du#

#={ac}/b^2(u-ln|u+1|)+C#

Now that the integration is done, express the result in terms of #x#. Substitute #u=b/cx# and get

#={ac}/b^2(b/cx-ln|b/cx+1|)+C#

Answer 2 · 2015-05-13T15:46:12

I have another method

Substitute #u = e^(7x)#

#du = 7e^(7x)dx#

#int(e^(7x))/(e^(14x)+9)dx = 1/7int(7e^(7x))/(e^(14x)+9)dx#

#u^2=e^(14x)#

#1/7int1/(u^2+9)du = 1/63int1/(1/9u^2+1)du#

Substitute again, #t = 1/3u# #t^2=1/9u^2#

#1/21int1/(t^2+1)dt#

#1/21[arctan(t)]+C#

Substitute back

#1/21arctan(1/3e^7x)+C#