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

Sunday, 27 September 2009

$\displaystyle\int\frac{{1}}{{{1}+\sqrt{{{2}{x}}}}}{\left.{d}{x}\right.}$ 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(u) du $\displaystyle{w}{h}{e}{r}{e}{w}{e}\le{t}$ u = g(x) $\displaystyle.$

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 $\displaystyle{d}{u}=$
1/sqrt(2x)dx $\displaystyle{b}{e}{c}{o}{m}{e}{s}:$

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

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

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

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

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

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

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

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|1+sqrt(2x) |+C

Help Notes

Sunday, 27 September 2009

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

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