Blog: `int_1^2 (ln(x))^2/x^3 dx` Evaluate the integral

$\displaystyle{\int_{{1}}^{{2}}}\frac{{{\left({\ln{{\left({x}\right)}}}\right)}}^{{2}}}{{{x}}^{{3}}}{\left.{d}{x}\right.}$

If f(x) and g(x) are differentiable function, then

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

If we rewrite f(x)=u and g'(x)=v, them

$\displaystyle\int{u}{v}{\left.{d}{x}\right.}={u}\int{v}{\left.{d}{x}\right.}-\int{\left({u}'\int{v}{\left.{d}{x}\right.}\right)}{\left.{d}{x}\right.}$

Using the above method of integration by parts,

Let's first evaluate the indefinite integral,

$\displaystyle\int\frac{{{\left({\ln{{\left({x}\right)}}}\right)}}^{{2}}}{{{x}}^{{3}}}{\left.{d}{x}\right.}={{\left({\ln{{\left({x}\right)}}}\right)}}^{{2}}\int\frac{{1}}{{{x}}^{{3}}}{\left.{d}{x}\right.}-\int{\left(\frac{{d}}{{\left.{d}{x}}}{{\left({\ln{{\left({x}\right)}}}\right)}}^{{2}}\int{\left(\frac{{1}}{{{x}}^{{3}}}\right)}{\left.{d}{x}\right.}\right)}{\left.{d}{x}\right.}$

$\displaystyle={{\left({\ln{{\left({x}\right)}}}\right)}}^{{2}}{\left(\frac{{{x}}^{{-{3}+{1}}}}{{-{3}+{1}}}\right)}-\int{\left({2}{\ln{{\left({x}\right)}}}{\left(\frac{{1}}{{x}}\right)}{\left(\frac{{{x}}^{{-{3}+{1}}}}{{-{3}+{1}}}\right)}\right)}{\left.{d}{x}\right.}$

$\displaystyle={{\left({\ln{{\left({x}\right)}}}\right)}}^{{2}}{\left(\frac{{{x}}^{{-{2}}}}{{-{{2}}}}\right)}-\int{\left({2}{\ln{{\left({x}\right)}}}{\left(\frac{{1}}{{x}}\right)}\frac{{{{x}}^{{-{2}}}}}{{-{{2}}}}\right)}{\left.{d}{x}\right.}$

$\displaystyle=-\frac{{1}}{{{2}{{x}}^{{2}}}}{{\left({\ln{{\left({x}\right)}}}\right)}}^{{2}}+\int{\left(\frac{{\ln{{\left({x}\right)}}}}{{{x}}^{{3}}}\right)}{\left.{d}{x}\right.}$

again applying integration by parts

$\displaystyle=-\frac{{1}}{{{2}{{x}}^{{2}}}}{{\left({\ln{{\left({x}\right)}}}\right)}}^{{2}}+{\ln{{\left({x}\right)}}}\cdot\int{\left(\frac{{1}}{{{x}}^{{3}}}\right)}{\left.{d}{x}\right.}-\int{\left(\frac{{d}}{{\left.{d}{x}}}{\left({\ln{{\left({x}\right)}}}\int{\left(\frac{{1}}{{{x}}^{{3}}}\right)}{\left.{d}{x}\right.}\right)}{\left.{d}{x}\right.}\right.}$

$\displaystyle=-\frac{{1}}{{{2}{{x}}^{{2}}}}{{\left({\ln{{\left({x}\right)}}}\right)}}^{{2}}+{\ln{{\left({x}\right)}}}\cdot{\left(\frac{{{x}}^{{-{3}+{1}}}}{{-{3}+{1}}}\right)}-\int{\left(\frac{{1}}{{x}}\right)}{\left(\frac{{{{x}}^{{-{{3}}}}+{1}}}{{-{3}+{1}}}\right)}{\left.{d}{x}\right.}$

$\displaystyle=-\frac{{1}}{{{2}{{x}}^{{2}}}}{{\left({\ln{{\left({x}\right)}}}\right)}}^{{2}}+{\ln{{\left({x}\right)}}}\cdot{\left(\frac{{{x}}^{{-{2}}}}{{-{{2}}}}\right)}-\int{\left(\frac{{1}}{{-{2}{{x}}^{{3}}}}\right)}{\left.{d}{x}\right.}$

$\displaystyle=-\frac{{1}}{{{2}{{x}}^{{2}}}}{{\left({\ln{{\left({x}\right)}}}\right)}}^{{2}}-\frac{{\ln{{\left({x}\right)}}}}{{{2}{{x}}^{{2}}}}+\frac{{1}}{{2}}\int{\left(\frac{{1}}{{{x}}^{{3}}}\right)}{\left.{d}{x}\right.}$

$\displaystyle=-\frac{{1}}{{{2}{{x}}^{{2}}}}{{\left({\ln{{\left({x}\right)}}}\right)}}^{{2}}-\frac{{\ln{{\left({x}\right)}}}}{{{2}{{x}}^{{2}}}}+\frac{{1}}{{2}}{\left(\frac{{{x}}^{{-{3}+{1}}}}{{-{3}+{1}}}\right)}$

$\displaystyle=-\frac{{1}}{{{2}{{x}}^{{2}}}}{{\left({\ln{{\left({x}\right)}}}\right)}}^{{2}}-\frac{{\ln{{\left({x}\right)}}}}{{{2}{{x}}^{{2}}}}-\frac{{1}}{{{4}{{x}}^{{2}}}}$

add a constant C to the solution,

$\displaystyle=-\frac{{1}}{{{2}{{x}}^{{2}}}}{\left({{\left({\ln{{\left({x}\right)}}}\right)}}^{{2}}+{\ln{{\left({x}\right)}}}+\frac{{1}}{{2}}\right)}+{C}$

Now let's evaluate definite integral,

$\displaystyle{\int_{{1}}^{{2}}}\frac{{{\left({\ln{{\left({x}\right)}}}\right)}}^{{2}}}{{{x}}^{{3}}}{\left.{d}{x}\right.}={{\left[-\frac{{1}}{{{2}{{x}}^{{2}}}}{\left({{\left({\ln{{\left({x}\right)}}}\right)}}^{{2}}+{\ln{{\left({x}\right)}}}+\frac{{1}}{{2}}\right)}\right]}_{{1}}^{{2}}}$

$\displaystyle={\left[-\frac{{1}}{{{2}\cdot{{2}}^{{2}}}}{\left({{\left({\ln{{\left({2}\right)}}}\right)}}^{{2}}+{\ln{{\left({2}\right)}}}+\frac{{1}}{{2}}\right)}\right]}-{\left[-\frac{{1}}{{{2}\cdot{{1}}^{{2}}}}{\left({{\left({\ln{{\left({1}\right)}}}\right)}}^{{2}}+{\ln{{\left({1}\right)}}}+\frac{{1}}{{2}}\right)}\right]}$

$\displaystyle={\left[-\frac{{1}}{{8}}{{\left({\ln{{\left({2}\right)}}}\right)}}^{{2}}-\frac{{\ln{{\left({2}\right)}}}}{{8}}-\frac{{1}}{{16}}+\frac{{1}}{{4}}\right]}$

$\displaystyle=\frac{{3}}{{16}}-\frac{{{\left({\ln{{\left({2}\right)}}}\right)}}^{{2}}}{{8}}-\frac{{\ln{{\left({2}\right)}}}}{{8}}$