Blog: `int (x e^(2x))/(1 + 2x)^2 dx` Evaluate the integral

Monday, November 14, 2016

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

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

If f(x) and g(x) are differentiable functions, then

$\displaystyle\int{f{{\left({x}\right)}}}{g{'}}{\left({x}\right)}={f{{\left({x}\right)}}}{g{{\left({x}\right)}}}-\int{f{'}}{\left({x}\right)}{g{{\left({x}\right)}}}{\left.{d}{x}\right.}$

If we rewrite f(x)=u and g'(x)=v, then

$\displaystyle\int{u}{v}{\left.{d}{x}\right.}={u}\int{v}{\left.{d}{x}\right.}-\int{\left({u}'\int{v}{\left.{d}{x}\right.}\right)}{\left.{d}{x}\right.}$

Using the above method of integration by parts,

Let $\displaystyle{u}={x}{{e}}^{{{2}{x}}}$

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

$\displaystyle{u}'={x}{\left({2}{{e}}^{{{2}{x}}}\right)}+{{e}}^{{{2}{x}}}$

$\displaystyle{u}'={{e}}^{{{2}{x}}}{\left({2}{x}+{1}\right)}$

$\displaystyle{v}=\frac{{1}}{{{\left({1}+{2}{x}\right)}}^{{2}}}$

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

Let's integrate by the use of substitution method,

Let t=1+2x

$\displaystyle{\left.{d}{t}\right.}={2}{\left.{d}{x}\right.}$

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

$\displaystyle=\frac{{1}}{{2}}{\left(\frac{{{t}}^{{-{2}+{1}}}}{{-{2}+{1}}}\right)}$

$\displaystyle=-\frac{{1}}{{{2}{t}}}$

substitute back t=1+2x,

$\displaystyle=-\frac{{1}}{{{2}{\left({1}+{2}{x}\right)}}}$

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

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

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

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

$\displaystyle=\frac{{{e}}^{{{2}{x}}}}{{4}}-\frac{{{x}{{e}}^{{{2}{x}}}}}{{{2}{\left({1}+{2}{x}\right)}}}$

Add a constant C to the solution,

$\displaystyle\int\frac{{{x}{{e}}^{{{2}{x}}}}}{{{\left({1}+{2}{x}\right)}}^{{2}}}{\left.{d}{x}\right.}=\frac{{{e}}^{{{2}{x}}}}{{4}}-\frac{{{x}{{e}}^{{{2}{x}}}}}{{{2}{\left({1}+{2}{x}\right)}}}+{C}$

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

If f(x) and g(x) are differentiable functions, then

$\displaystyle\int{f{{\left({x}\right)}}}{g{'}}{\left({x}\right)}={f{{\left({x}\right)}}}{g{{\left({x}\right)}}}-\int{f{'}}{\left({x}\right)}{g{{\left({x}\right)}}}{\left.{d}{x}\right.}$

If we rewrite f(x)=u and g'(x)=v, then

$\displaystyle\int{u}{v}{\left.{d}{x}\right.}={u}\int{v}{\left.{d}{x}\right.}-\int{\left({u}'\int{v}{\left.{d}{x}\right.}\right)}{\left.{d}{x}\right.}$

Using the above method of integration by parts,

Let $\displaystyle{u}={x}{{e}}^{{{2}{x}}}$

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

$\displaystyle{u}'={x}{\left({2}{{e}}^{{{2}{x}}}\right)}+{{e}}^{{{2}{x}}}$

$\displaystyle{u}'={{e}}^{{{2}{x}}}{\left({2}{x}+{1}\right)}$

$\displaystyle{v}=\frac{{1}}{{{\left({1}+{2}{x}\right)}}^{{2}}}$

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

Let's integrate by the use of substitution method,

Let t=1+2x

$\displaystyle{\left.{d}{t}\right.}={2}{\left.{d}{x}\right.}$

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

$\displaystyle=\frac{{1}}{{2}}{\left(\frac{{{t}}^{{-{2}+{1}}}}{{-{2}+{1}}}\right)}$

$\displaystyle=-\frac{{1}}{{{2}{t}}}$

substitute back t=1+2x,

$\displaystyle=-\frac{{1}}{{{2}{\left({1}+{2}{x}\right)}}}$

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

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

$\displaystyle=\frac{{{e}}^{{{2}{x}}}}{{4}}-\frac{{{x}{{e}}^{{{2}{x}}}}}{{{2}{\left({1}+{2}{x}\right)}}}$

Add a constant C to the solution,

$\displaystyle\int\frac{{{x}{{e}}^{{{2}{x}}}}}{{{\left({1}+{2}{x}\right)}}^{{2}}}{\left.{d}{x}\right.}=\frac{{{e}}^{{{2}{x}}}}{{4}}-\frac{{{x}{{e}}^{{{2}{x}}}}}{{{2}{\left({1}+{2}{x}\right)}}}+{C}$

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Monday, November 14, 2016

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

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What is the Exposition, Rising Action, Climax, and Falling Action of "One Thousand Dollars"?

Monday, November 14, 2016

Evaluate the integral

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What is the Exposition, Rising Action, Climax, and Falling Action of &quot;One Thousand Dollars&quot;?

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

What is the Exposition, Rising Action, Climax, and Falling Action of "One Thousand Dollars"?