`int 1/(1 + sqrt(2x)) dx` Find the indefinite integral by u substitution. (let u be the denominator of the integral)

Thursday, November 13, 2014

$\displaystyle\int\frac{{1}}{{{1}+\sqrt{{{2}{x}}}}}{\left.{d}{x}\right.}$ Find the indefinite integral by u substitution. (let u be the denominator of the integral)

Solving for indefinite integral using u-substitution follows:

$\displaystyle\int{f{{\left({g{{\left({x}\right)}}}\right)}}}\cdot{g{'}}{\left({x}\right)}{\left.{d}{x}\right.}=\int{f{{\left({u}\right)}}}{d}{u}$ where we let $\displaystyle{u}={g{{\left({x}\right)}}}$ .

In this case, it is stated that to let u be the denominator of integral which means let:

$\displaystyle{u}={1}+\sqrt{{{2}{x}}}.$

This can be rearrange into $\displaystyle\sqrt{{{2}{x}}}={u}-{1}$

Finding the derivative of u : $\displaystyle{d}{u}=\frac{{1}}{\sqrt{{{2}{x}}}}{\left.{d}{x}\right.}$

Substituting $\displaystyle\sqrt{{{2}{x}}}={u}-{1}$ into $\displaystyle{d}{u}=\frac{{1}}{\sqrt{{{2}{x}}}}{\left.{d}{x}\right.}$ becomes:

$\displaystyle{d}{u}=\frac{{1}}{{{u}-{1}}}{\left.{d}{x}\right.}$

Rearranged into $\displaystyle{\left({u}-{1}\right)}{d}{u}={\left.{d}{x}\right.}$

Applying u-substitution using $\displaystyle{u}={1}+\sqrt{{{2}{x}}}$ and $\displaystyle{\left({u}-{1}\right)}{d}{u}={d}{u}$ :

$\displaystyle\int\frac{{1}}{{{1}+\sqrt{{{2}{x}}}}}{\left.{d}{x}\right.}=\int\frac{{{u}-{1}}}{{u}}\cdot{d}{u}$

Express into two separate fractions:

$\displaystyle\int\frac{{{u}-{1}}}{{u}}\cdot{d}{u}=\int{\left(\frac{{u}}{{u}}-\frac{{1}}{{u}}\right)}{d}{u}$

$\displaystyle=\int{\left({1}-\frac{{1}}{{u}}\right)}{d}{u}$

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

$\displaystyle\int{\left({1}-\frac{{1}}{{u}}\right)}{d}{u}=\int{1}{d}{u}-\int\frac{{1}}{{u}}{d}{u}$

$\displaystyle={u}-{\ln}{\left|{u}\right|}+{C}$

Substitute $\displaystyle{u}={1}+\sqrt{{{2}{x}}}$ to the $\displaystyle{u}-{\ln}{\left|{u}\right|}+{C}$ :

$\displaystyle{u}-{\ln}{\left|{u}\right|}+{C}={1}+\sqrt{{{2}{x}}}-{\ln}{\left|{1}+\sqrt{{{2}{x}}}\right|}+{C}$

Ninenotes

Thursday, November 13, 2014

$\displaystyle\int\frac{{1}}{{{1}+\sqrt{{{2}{x}}}}}{\left.{d}{x}\right.}$ Find the indefinite integral by u substitution. (let u be the denominator of the integral)

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