`int 3/(2sqrt(x)(1+x)) dx` Find the indefinite integral

Friday, October 7, 2011

$\displaystyle\int\frac{{3}}{{{2}\sqrt{{{x}}}{\left({1}+{x}\right)}}}{\left.{d}{x}\right.}$ Find the indefinite integral

For the given integral: $\displaystyle\int\frac{{3}}{{{2}\sqrt{{{x}}}{\left({1}+{x}\right)}}}{\left.{d}{x}\right.}$ , we may apply the basic integration property: $\displaystyle\int{c}\cdot{f{{\left({x}\right)}}}{\left.{d}{x}\right.}={c}\int{f{{\left({x}\right)}}}{\left.{d}{x}\right.}$ .

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

For the integral part, we apply u-substitution by letting:

$\displaystyle{u}=\sqrt{{{x}}}$

We square both sides to get: $\displaystyle{{u}}^{{2}}={x}$ .

Then apply implicit differentiation, we take the derivative on both sides with respect to x as:

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

Plug-in $\displaystyle{\left.{d}{x}\right.}={2}{u}{d}{u}$ , $\displaystyle{u}=\sqrt{{{x}}}$ and $\displaystyle{x}={{u}}^{{2}}$ in the integral:

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

Simplify by cancelling out u and 2 from top and bottom:

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

The integral part resembles the basic integration formula for inverse tangent:

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

then,

$\displaystyle{3}\int\frac{{1}}{{{1}+{{u}}^{{2}}}}{d}{u}={3}\cdot{\arctan{{\left({u}\right)}}}+{C}$

Express in terms x by plug-in $\displaystyle{u}=\sqrt{{{x}}}$ :

$\displaystyle{3}{\arctan{{\left({u}\right)}}}+{C}={3}{\arctan{{\left(\sqrt{{{x}}}\right)}}}+{C}$

Final answer:

$\displaystyle\int\frac{{3}}{{{2}\sqrt{{{x}}}{\left({1}+{x}\right)}}}{\left.{d}{x}\right.}={3}{\arctan{{\left(\sqrt{{{x}}}\right)}}}+{C}$

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Friday, October 7, 2011

$\displaystyle\int\frac{{3}}{{{2}\sqrt{{{x}}}{\left({1}+{x}\right)}}}{\left.{d}{x}\right.}$ Find the indefinite 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?

Friday, October 7, 2011

Find the indefinite integral

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Post a Comment

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

$\displaystyle\int\frac{{3}}{{{2}\sqrt{{{x}}}{\left({1}+{x}\right)}}}{\left.{d}{x}\right.}$ Find the indefinite integral

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