Integral from 0 to ln 3 of $\frac{e^{3 x}}{e^{6 x} + 5}$ $e^(3x)/(e^(6x)+5)$ . Help please?

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Answer 1 · 2018-05-02T08:58:36

See process below

$I = \int \frac{e^{3 x}}{e^{6 x} + 5} d x$ $I=inte^(3x)/(e^(6x)+5)dx$

Change of variable $t = e^{3 x}$ $t=e^(3x)$ , then $d t = 3 e^{3 x} d x = 3 t d x$ $dt=3e^(3x)dx=3tdx$

With this change we have

$I=intcancelt/(t^2+5)·1/(3cancelt)dt=1/3int1/(t^2+5)$

Other change of variable $t=sqrt5tantheta$ . Then $dt=sqrt5sec^2theta$ and integral

$I=1/3int1/(5tan^2theta+5)·sqrt5sec^2thetad theta=$

$=1/3·sqrt5/5int(cancelsec^2thetad theta)/(cancelsec^2theta)=$

$=sqrt5/15intd theta=sqrt5/15theta=sqrt5/15arctan(t/sqrt5)=$

$=sqrt5/15arctan(e^(3x)/sqrt5)=F(x)$

$F(ln3)-F(0)=sqrt5/15arctan(27/sqrt5)-sqrt5/15arctan(1/sqrt5)$

Integral from 0 to ln 3 of e^(3x)/(e^(6x)+5)e3xe6x+5e^(3x)/(e^(6x)+5) . Help please?