Notes: `int_e^(e^4) (dx)/(xsqrt(ln(x)))` Evaluate the definite integral.

Wednesday, 24 February 2016

$\displaystyle{\int_{{e}}^{{{{e}}^{{4}}}}}\frac{{{\left.{d}{x}\right.}}}{{{x}\sqrt{{{\ln{{\left({x}\right)}}}}}}}$ Evaluate the definite integral.

You need to use the following substitution ln x=u, such that:

$\displaystyle{\ln{{x}}}={u}\Rightarrow\frac{{{\left.{d}{x}\right.}}}{{x}}={d}{u}$

$\displaystyle{\int_{{e}}^{{{{e}}^{{4}}}}}\frac{{{\left.{d}{x}\right.}}}{{{x}\cdot\sqrt{{{\ln{{x}}}}}}}={\int_{{{u}_{{1}}}}^{{{u}_{{2}}}}}\frac{{{d}{u}}}{{\sqrt{{u}}}}$

$\displaystyle{\int_{{{u}_{{1}}}}^{{{u}_{{2}}}}}\frac{{{d}{u}}}{{\sqrt{{u}}}}={2}\sqrt{{u}}{{\mid}_{{{u}_{{1}}}}^{{{u}_{{2}}}}}$

Replacing back ln x for u yields:

$\displaystyle{\int_{{e}}^{{{{e}}^{{4}}}}}\frac{{{\left.{d}{x}\right.}}}{{{x}\cdot\sqrt{{{\ln{{x}}}}}}}={2}\sqrt{{{\ln{{x}}}}}{{\mid}_{{e}}^{{{{e}}^{{4}}}}}$

Using Leibniz-Newton theorem yields:

$\displaystyle{\int_{{e}}^{{{{e}}^{{4}}}}}\frac{{{\left.{d}{x}\right.}}}{{{x}\cdot\sqrt{{{\ln{{x}}}}}}}={2}\sqrt{{{{\ln{{e}}}}^{{4}}}}-{2}\sqrt{{{\ln{{e}}}}}$

$\displaystyle{\int_{{e}}^{{{{e}}^{{4}}}}}\frac{{{\left.{d}{x}\right.}}}{{{x}\cdot\sqrt{{{\ln{{x}}}}}}}={2}\sqrt{{4}}-{2}\sqrt{{1}}$

$\displaystyle{\int_{{e}}^{{{{e}}^{{4}}}}}\frac{{{\left.{d}{x}\right.}}}{{{x}\cdot\sqrt{{{\ln{{x}}}}}}}={4}-{2}$

$\displaystyle{\int_{{e}}^{{{{e}}^{{4}}}}}\frac{{{\left.{d}{x}\right.}}}{{{x}\cdot\sqrt{{{\ln{\ldots}}}}}}$