Blog: `int ln(root(3)(x)) dx` Evaluate the integral

Tuesday, August 4, 2015

$\displaystyle\int{\ln{{\left({\sqrt[{{3}}]{{{x}}}}\right)}}}{\left.{d}{x}\right.}$ Evaluate the integral

$\displaystyle\int{\ln{{\left({\sqrt[{{3}}]{{{x}}}}\right)}}}{\left.{d}{x}\right.}$

To evaluate, apply integration by parts $\displaystyle\int{u}{d}{v}={u}{v}-\int{v}{d}{u}$ .

So let

$\displaystyle{u}={\ln{{\left({\sqrt[{{3}}]{{{x}}}}\right)}}}={\ln{{\left({{x}}^{{\frac{{1}}{{3}}}}\right)}}}=\frac{{1}}{{3}}{\ln{{\left({x}\right)}}}$

and

$\displaystyle{d}{v}={\left.{d}{x}\right.}$

Then, differentiate u and integrate dv.

$\displaystyle{d}{u}=\frac{{1}}{{3}}\cdot\frac{{1}}{{x}}{\left.{d}{x}\right.}=\frac{{1}}{{{3}{x}}}{\left.{d}{x}\right.}$

and

$\displaystyle{v}=\int{\left.{d}{x}\right.}={x}$

And, plug-in them to the formula. So the integral becomes:

$\displaystyle\int{\ln{{\left({\sqrt[{{3}}]{{{x}}}}\right)}}}{\left.{d}{x}\right.}$

$\displaystyle=\frac{{1}}{{3}}{\ln{{\left({x}\right)}}}\cdot{x}-\int{x}\cdot\frac{{1}}{{{3}{x}}}{\left.{d}{x}\right.}$

$\displaystyle=\frac{{1}}{{3}}{x}{\ln{{\left({x}\right)}}}-\int\frac{{1}}{{3}}{\left.{d}{x}\right.}$

$\displaystyle=\frac{{1}}{{3}}{x}{\ln{{\left({x}\right)}}}-\frac{{1}}{{3}}{x}+{C}$

Therefore, $\displaystyle\int{\ln{{\left({\sqrt[{{3}}]{{{x}}}}\right)}}}{\left.{d}{x}\right.}=\frac{{1}}{{3}}{x}{\ln{{\left({x}\right)}}}-\frac{{1}}{{3}}{x}+{C}$ .

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Tuesday, August 4, 2015

$\displaystyle\int{\ln{{\left({\sqrt[{{3}}]{{{x}}}}\right)}}}{\left.{d}{x}\right.}$ Evaluate the integral

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