Blog: `int (sqrt(x)/(sqrt(x) - 3)) dx` Find the indefinite integral by u substitution. (let u be the denominator of the integral)

To apply u-substitution , we let $\displaystyle{u}=\sqrt{{{x}}}-{3}$ .

Then $\displaystyle{d}{u}=\frac{{1}}{{{2}\sqrt{{{x}}}{\left.{d}{x}\right.}}}$ .

Rearrange $\displaystyle{d}{u}=\frac{{1}}{{{2}\sqrt{{{x}}}}}{\left.{d}{x}\right.}$ into $\displaystyle{\left.{d}{x}\right.}={2}\sqrt{{{x}}}{d}{u}$

Substituting $\displaystyle{\left.{d}{x}\right.}={2}\sqrt{{{x}}}{d}{u}$ and $\displaystyle{u}=\sqrt{{{x}}}-{3}$ :

$\displaystyle\int\frac{\sqrt{{{x}}}}{{\sqrt{{{x}}}-{3}}}{\left.{d}{x}\right.}=\int\frac{\sqrt{{{x}}}}{{u}}\cdot{2}\sqrt{{{x}}}{\left.{d}{x}\right.}$

Simplify: $\displaystyle\sqrt{{{x}}}\cdot\sqrt{{{x}}}={x}$

$\displaystyle\int\frac{\sqrt{{{x}}}}{{u}}\cdot{2}\sqrt{{{x}}}{d}{u}=\int\frac{{{2}{x}}}{{u}}{d}{u}$

Rearrange $\displaystyle{u}=\sqrt{{{x}}}-{3}$ into $\displaystyle\sqrt{{{x}}}={u}+{3}$

Squaring both sides of $\displaystyle\sqrt{{{x}}}={u}+{3}$ then

$\displaystyle{x}={{u}}^{{2}}+{6}{u}+{9}$

$\displaystyle\int\frac{{{2}{x}}}{{u}}{d}{u}={2}\int\frac{{{{u}}^{{2}}+{6}{u}+{9}}}{{u}}{d}{u}$

...

To apply u-substitution , we let $\displaystyle{u}=\sqrt{{{x}}}-{3}$ .

Then $\displaystyle{d}{u}=\frac{{1}}{{{2}\sqrt{{{x}}}{\left.{d}{x}\right.}}}$ .

Rearrange $\displaystyle{d}{u}=\frac{{1}}{{{2}\sqrt{{{x}}}}}{\left.{d}{x}\right.}$ into $\displaystyle{\left.{d}{x}\right.}={2}\sqrt{{{x}}}{d}{u}$

Substituting $\displaystyle{\left.{d}{x}\right.}={2}\sqrt{{{x}}}{d}{u}$ and $\displaystyle{u}=\sqrt{{{x}}}-{3}$ :

$\displaystyle\int\frac{\sqrt{{{x}}}}{{\sqrt{{{x}}}-{3}}}{\left.{d}{x}\right.}=\int\frac{\sqrt{{{x}}}}{{u}}\cdot{2}\sqrt{{{x}}}{\left.{d}{x}\right.}$

Simplify: $\displaystyle\sqrt{{{x}}}\cdot\sqrt{{{x}}}={x}$

$\displaystyle\int\frac{\sqrt{{{x}}}}{{u}}\cdot{2}\sqrt{{{x}}}{d}{u}=\int\frac{{{2}{x}}}{{u}}{d}{u}$

Rearrange $\displaystyle{u}=\sqrt{{{x}}}-{3}$ into $\displaystyle\sqrt{{{x}}}={u}+{3}$

Squaring both sides of $\displaystyle\sqrt{{{x}}}={u}+{3}$ then

$\displaystyle{x}={{u}}^{{2}}+{6}{u}+{9}$

$\displaystyle\int\frac{{{2}{x}}}{{u}}{d}{u}={2}\int\frac{{{{u}}^{{2}}+{6}{u}+{9}}}{{u}}{d}{u}$

$\displaystyle={2}\int{\left(\frac{{{u}}^{{2}}}{{u}}+{6}\frac{{u}}{{u}}+\frac{{9}}{{u}}\right)}{d}{u}$

$\displaystyle={2}\int{\left({u}+{6}+\frac{{9}}{{u}}\right)}{d}{u}$

$\displaystyle={2}\cdot{\left(\frac{{{u}}^{{2}}}{{2}}+{6}{u}+{9}{\ln{{\text{abs}}}}{\left|{u}\right|}\right)}+{C}$

Substitute u =sqrt(x)-3:

$\displaystyle{2}\cdot{\left(\frac{{{u}}^{{2}}}{{2}}+{6}{u}+{9}{\ln}{\left|{u}\right|}\right)}+{C}={2}\cdot{\left(\frac{{{\left(\sqrt{{{x}}}-{3}\right)}}^{{2}}}{{2}}+{6}{\left(\sqrt{{{x}}}-{3}\right)}+{9}{\ln}{\left|{\left(\sqrt{{{x}}}-{3}\right)}\right|}\right)}+{C}$

$\displaystyle={{\left(\sqrt{{{x}}}-{3}\right)}}^{{2}}+{12}{\left(\sqrt{{{x}}}-{3}\right)}+{18}{\ln}{\left|{\left(\sqrt{{{x}}}-{3}\right)}\right|}+{C}$

$\displaystyle={x}-{6}\sqrt{{{x}}}+{9}+{12}\sqrt{{{x}}}-{36}+{18}{\ln}{\left|\sqrt{{{x}}}-{3}\right|}+{C}$

$\displaystyle={x}+{6}\sqrt{{{x}}}-{27}+{18}{\ln}{\left|\sqrt{{{x}}}-{3}\right|}+{C}$

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$\displaystyle\int{\left(\frac{\sqrt{{{x}}}}{{\sqrt{{{x}}}-{3}}}\right)}{\left.{d}{x}\right.}$ Find the indefinite integral by u substitution. (let u be the denominator of the integral)

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What are the problems with Uganda's government?

Find the indefinite integral by u substitution. (let u be the denominator of the integral)

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What are the problems with Uganda&#39;s government?

$\displaystyle\int{\left(\frac{\sqrt{{{x}}}}{{\sqrt{{{x}}}-{3}}}\right)}{\left.{d}{x}\right.}$ Find the indefinite integral by u substitution. (let u be the denominator of the integral)

What are the problems with Uganda's government?