New Notes Blog: `int 2 / (7e^x + 4) dx` Find the indefinite integral

Tuesday, 18 February 2014

$\displaystyle\int\frac{{2}}{{{7}{{e}}^{{x}}+{4}}}{\left.{d}{x}\right.}$ Find the indefinite integral

To evaluate the given integral problem: $\displaystyle\int\frac{{2}}{{{7}{{e}}^{{x}}+{4}}}{\left.{d}{x}\right.}$ , we may apply u-substitution using: $\displaystyle{u}={{e}}^{{x}}$ then $\displaystyle{d}{u}={{e}}^{{x}}{\left.{d}{x}\right.}$ .

Plug-in $\displaystyle{u}={{e}}^{{x}}$ on $\displaystyle{d}{u}={{e}}^{{x}}{\left.{d}{x}\right.}$ , we get: $\displaystyle{d}{u}={u}{\left.{d}{x}\right.}$ or $\displaystyle\frac{{{d}{u}}}{{u}}={\left.{d}{x}\right.}$

The integral becomes:

$\displaystyle\int\frac{{2}}{{{7}{{e}}^{{x}}+{4}}}{\left.{d}{x}\right.}=\int\frac{{2}}{{{7}{u}+{4}}}\cdot\frac{{{d}{u}}}{{u}}$

$\displaystyle=\int\frac{{2}}{{{7}{{u}}^{{2}}+{4}{u}}}{d}{u}$

Apply the basic properties of integration: $\displaystyle\int{c}\cdot{f{{\left({x}\right)}}}{\left.{d}{x}\right.}={c}\int{f{{\left({x}\right)}}}\ldots$

The integral becomes:

$\displaystyle\int\frac{{2}}{{{7}{{e}}^{{x}}+{4}}}{\left.{d}{x}\right.}=\int\frac{{2}}{{{7}{u}+{4}}}\cdot\frac{{{d}{u}}}{{u}}$

$\displaystyle=\int\frac{{2}}{{{7}{{u}}^{{2}}+{4}{u}}}{d}{u}$

Apply the basic properties of integration: $\displaystyle\int{c}\cdot{f{{\left({x}\right)}}}{\left.{d}{x}\right.}={c}\int{f{{\left({x}\right)}}}{\left.{d}{x}\right.}$ .

$\displaystyle\int\frac{{2}}{{{7}{{u}}^{{2}}+{4}{u}}}{d}{u}={2}\int\frac{{1}}{{{7}{{u}}^{{2}}+{4}{u}}}{d}{u}$

Apply completing the square: $\displaystyle{7}{{u}}^{{2}}+{4}{u}={{\left(\sqrt{{{7}}}{u}+\frac{{2}}{\sqrt{{{7}}}}\right)}}^{{2}}-\frac{{4}}{{7}}$

$\displaystyle{2}\int\frac{{1}}{{{7}{{u}}^{{2}}+{4}{u}}}{d}{u}={2}\int\frac{{1}}{{{{\left(\sqrt{{{7}}}{u}+\frac{{2}}{\sqrt{{{7}}}}\right)}}^{{2}}-\frac{{4}}{{7}}}}{d}{u}$

Let $\displaystyle{v}=\sqrt{{{7}}}{u}+\frac{{2}}{\sqrt{{{7}}}}$ then $\displaystyle{d}{v}=\sqrt{{{7}}}{d}{u}$ or $\displaystyle\frac{{{d}{v}}}{\sqrt{{{7}}}}={d}{u}$ .

The integral becomes:

$\displaystyle{2}\int\frac{{1}}{{{7}{{u}}^{{2}}+{4}{u}}}{d}{u}={2}\int\frac{{1}}{{{{v}}^{{2}}-\frac{{4}}{{7}}}}\cdot\frac{{{d}{v}}}{\sqrt{{{7}}}}$

Rationalize the denominator:

$\displaystyle{2}\int\frac{{1}}{{{{v}}^{{2}}-\frac{{4}}{{7}}}}\cdot\frac{{{d}{v}}}{\sqrt{{{7}}}}\cdot\frac{\sqrt{{{7}}}}{\sqrt{{{7}}}}$

$\displaystyle={2}\int\frac{{\sqrt{{{7}}}{d}{v}}}{{{7}\cdot{\left({{v}}^{{2}}-\frac{{4}}{{7}}\right)}}}$

$\displaystyle={2}\int\frac{{\sqrt{{{7}}}{d}{v}}}{{{7}{{v}}^{{2}}-{4}}}$

From the table of integrals, we may apply $\displaystyle\int\frac{{\left.{d}{x}}}{{{{x}}^{{2}}-{{a}}^{{2}}}}=\frac{{1}}{{{2}{a}}}{\ln{{\left[\frac{{{u}-{a}}}{{{u}+{a}}}\right]}}}+{C}$

Let: $\displaystyle{w}=\sqrt{{{7}}}{v}$ then $\displaystyle{d}{w}=\sqrt{{{7}}}{d}{v}$

$\displaystyle{2}\int\frac{{\sqrt{{{7}}}{d}{v}}}{{{7}{{v}}^{{2}}-{4}}}={2}\int\frac{{\sqrt{{{7}}}{d}{v}}}{{{{\left(\sqrt{{{7}}}{v}\right)}}^{{2}}-{{2}}^{{2}}}}$

$\displaystyle={2}\int\frac{{{d}{w}}}{{{{w}}^{{2}}-{{2}}^{{2}}}}$

$\displaystyle={2}\cdot\frac{{1}}{{{2}\cdot{2}}}{\ln{{\left[\frac{{{w}-{2}}}{{{w}+{2}}}\right]}}}+{C}$

$\displaystyle=\frac{{1}}{{2}}{\ln{{\left[\frac{{{w}-{2}}}{{{w}+{2}}}\right]}}}+{C}$

Recall we let: $\displaystyle{w}=\sqrt{{{7}}}{v}$ and $\displaystyle{v}=\sqrt{{{7}}}{u}+\frac{{2}}{\sqrt{{{7}}}}$ .

Then, $\displaystyle{w}=\sqrt{{{7}}}\cdot{\left[\sqrt{{{7}}}{u}+\frac{{2}}{\sqrt{{{7}}}}\right]}={7}{u}+{2}$

Plug-in $\displaystyle{u}={{e}}^{{x}}$ on $\displaystyle{w}={7}{u}+{2}$ , we get: $\displaystyle{w}={7}{{e}}^{{x}}+{2}$

The indefinite integral will be:

$\displaystyle\int\frac{{2}}{{{7}{{e}}^{{x}}+{4}}}{\left.{d}{x}\right.}=\frac{{1}}{{2}}{\ln{{\left[\frac{{{7}{{e}}^{{x}}+{2}-{2}}}{{{7}{{e}}^{{x}}+{2}+{2}}}\right]}}}+{C}$

$\displaystyle=\frac{{1}}{{2}}{\ln{{\left[\frac{{{7}{{e}}^{{x}}}}{{{7}{{e}}^{{x}}+{4}}}\right]}}}+{C}$

New Notes Blog

Tuesday, 18 February 2014

$\displaystyle\int\frac{{2}}{{{7}{{e}}^{{x}}+{4}}}{\left.{d}{x}\right.}$ Find the indefinite integral

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In "By the Waters of Babylon," under the leadership of John, what do you think the Hill People will do with their society?

Tuesday, 18 February 2014

Find the indefinite integral

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Post a Comment

In &quot;By the Waters of Babylon,&quot; under the leadership of John, what do you think the Hill People will do with their society?

$\displaystyle\int\frac{{2}}{{{7}{{e}}^{{x}}+{4}}}{\left.{d}{x}\right.}$ Find the indefinite integral

In "By the Waters of Babylon," under the leadership of John, what do you think the Hill People will do with their society?