`int` `e^(3x)/(e^x+e^(3x)) dx`

Friday, June 5, 2015

$\displaystyle\int$ $\displaystyle\frac{{{e}}^{{{3}{x}}}}{{{{e}}^{{x}}+{{e}}^{{{3}{x}}}}}{\left.{d}{x}\right.}$

$\displaystyle\int\frac{{{e}}^{{{3}{x}}}}{{{{e}}^{{x}}+{{e}}^{{{3}{x}}}}}{\left.{d}{x}\right.}$

To solve this, let's simplify first the integrand.

$\displaystyle=\int\frac{{{e}}^{{{3}{x}}}}{{{{e}}^{{x}}{\left({1}+{{e}}^{{{2}{x}}}\right)}}}{\left.{d}{x}\right.}$

$\displaystyle=\int\frac{{{{e}}^{{x}}\cdot{{e}}^{{{2}{x}}}}}{{{{e}}^{{x}}{\left({1}+{{e}}^{{{2}{x}}}\right)}}}{\left.{d}{x}\right.}$

$\displaystyle=\int\frac{{{e}}^{{{2}{x}}}}{{{1}+{{e}}^{{{2}{x}}}}}{\left.{d}{x}\right.}$

Then, apply u-substitution method.

$\displaystyle{u}={1}+{{e}}^{{{2}{x}}}$

$\displaystyle{d}{u}={{e}}^{{{2}{x}}}\cdot{2}{\left.{d}{x}\right.}$

$\displaystyle\frac{{{d}{u}}}{{2}}={{e}}^{{{2}{x}}}{\left.{d}{x}\right.}$

Expressing the integral in terms of u, it becomes:

$\displaystyle=\int\frac{{1}}{{{1}+{{e}}^{{{2}{x}}}}}\cdot{{e}}^{{{2}{x}}}{\left.{d}{x}\right.}$

$\displaystyle=\int\frac{{1}}{{u}}\cdot\frac{{{d}{u}}}{{2}}$

$\displaystyle=\frac{{1}}{{2}}\int\frac{{1}}{{u}}{d}{u}$

$\displaystyle=\frac{{1}}{{2}}{\ln}{\left|{u}\right|}+{C}$

And, substitute back $\displaystyle{u}={1}+{{e}}^{{{2}{x}}}$ .

$\displaystyle=\frac{{1}}{{2}}{\ln}{\left|{1}+{{e}}^{{{2}{x}}}\right|}+{C}$

Therefore, $\displaystyle\int\frac{{{e}}^{{{3}{x}}}}{{{{e}}^{{x}}+{{e}}^{{{3}{x}}}}}{\left.{d}{x}\right.}=\frac{{1}}{{2}}{\ln}{\left|{1}+{{e}}^{{{2}{x}}}\right|}+{C}$ .

Ninenotes

Friday, June 5, 2015

$\displaystyle\int$ $\displaystyle\frac{{{e}}^{{{3}{x}}}}{{{{e}}^{{x}}+{{e}}^{{{3}{x}}}}}{\left.{d}{x}\right.}$

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