`int_0^ln5 e^x/(1+e^(2x)) dx` Evaluate the definite integral

Thursday, October 8, 2009

$\displaystyle{\int_{{0}}^{{\ln{{5}}}}}\frac{{{e}}^{{x}}}{{{1}+{{e}}^{{{2}{x}}}}}{\left.{d}{x}\right.}$ Evaluate the definite integral

For the given integral problem: $\displaystyle{\int_{{0}}^{{{\ln{{\left({5}\right)}}}}}}\frac{{{e}}^{{x}}}{{{1}+{{e}}^{{{2}{x}}}}}{\left.{d}{x}\right.}$ , it resembles the basic integration formula for inverse tangent:

$\displaystyle{\int_{{a}}^{{b}}}\frac{{{d}{u}}}{{{{u}}^{{2}}+{{c}}^{{2}}}}={\left(\frac{{1}}{{c}}\right)}{\arctan{{\left(\frac{{u}}{{c}}\right)}}}{{\mid}_{{a}}^{{b}}}$

where we let:

$\displaystyle{{u}}^{{2}}={{e}}^{{{2}{x}}}$ or $\displaystyle{{\left({{e}}^{{x}}\right)}}^{{2}}$ then $\displaystyle{u}={{e}}^{{x}}$

$\displaystyle{{c}}^{{2}}={1}$ or $\displaystyle{{1}}^{{2}}$ then $\displaystyle{c}={1}$

For the derivative of $\displaystyle{u}={{e}}^{{{x}}}$ , we apply the derivative of exponential function:

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

Applying u-substitution: $\displaystyle{u}={{e}}^{{x}}$ and $\displaystyle{d}{u}={{e}}^{{x}}{\left.{d}{x}\right.}$ , we get:

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

$\displaystyle=\int\frac{{{d}{u}}}{{{1}+{{\left({u}\right)}}^{{2}}}}$

Applying the basic integral formula of inverse tangent, we get:

$\displaystyle\int\frac{{{d}{u}}}{{{1}+{{\left({u}\right)}}^{{2}}}}={\left(\frac{{1}}{{1}}\right)}{\arctan{{\left(\frac{{u}}{{1}}\right)}}}$

= $\displaystyle{\arctan{{\left({u}\right)}}}$

Express it in terms of x by plug-in $\displaystyle{u}={{e}}^{{x}}$ :

$\displaystyle{\arctan{{\left({u}\right)}}}={\arctan{{\left({{e}}^{{x}}\right)}}}$

Evaluate with the given boundary limit:

$\displaystyle{\arctan{{\left({{e}}^{{x}}\right)}}}{{\mid}_{{0}}^{{{\ln{{\left({5}\right)}}}}}}={\arctan{{\left({{e}}^{{{\ln{{\left({5}\right)}}}}}\right)}}}-{\arctan{{\left({{e}}^{{0}}\right)}}}$

$\displaystyle={\arctan{{\left({5}\right)}}}-{\arctan{{\left({1}\right)}}}$

$\displaystyle={\arctan{{\left({5}\right)}}}-\frac{\pi}{{4}}$

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Thursday, October 8, 2009

$\displaystyle{\int_{{0}}^{{\ln{{5}}}}}\frac{{{e}}^{{x}}}{{{1}+{{e}}^{{{2}{x}}}}}{\left.{d}{x}\right.}$ Evaluate the definite integral

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Thomas Jefferson's election in 1800 is sometimes called the Revolution of 1800. Why could it be described in this way?

Thursday, October 8, 2009

Evaluate the definite integral

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Thomas Jefferson&#39;s election in 1800 is sometimes called the Revolution of 1800. Why could it be described in this way?

$\displaystyle{\int_{{0}}^{{\ln{{5}}}}}\frac{{{e}}^{{x}}}{{{1}+{{e}}^{{{2}{x}}}}}{\left.{d}{x}\right.}$ Evaluate the definite integral

Thomas Jefferson's election in 1800 is sometimes called the Revolution of 1800. Why could it be described in this way?