What is the integral from 1 to 5 of (ln(x^2))/(x^3) ?

The integral from 1 to 5 of (ln(x^2))/(x^3) is 12/25-1/25 ln(5)

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integral from 1 to 5 of (ln(x^2))/(x^3)

\int_{1}^{5} \frac{ln ( x ^{2} )}{x ^{3}} d x

\frac{12}{25} - \frac{1}{25} ln (5)

Decimal Notation

0.41562 \dots

Solution steps

\int_{1}^{5} \frac{ln ( x ^{2} )}{x ^{3}} d x

Apply Integration By Parts

: [- \frac{ln ( x ^{2} )}{2 x ^{2}} - \int - \frac{1}{x ^{3}} d x]_{1}^{5}

= [- \frac{ln ( x ^{2} )}{2 x ^{2}} - \int - \frac{1}{x ^{3}} d x]_{1}^{5}

\int - \frac{1}{x ^{3}} d x = \frac{1}{2 x ^{2}}

= [- \frac{ln ( x ^{2} )}{2 x ^{2}} - \frac{1}{2 x ^{2}}]_{1}^{5}

\frac{ln ( x ^{2} )}{2 x ^{2}} = \frac{ln ( x )}{x ^{2}}

= [- \frac{ln ( x )}{x ^{2}} - \frac{1}{2 x ^{2}}]_{1}^{5}

Compute the boundaries:

\frac{12}{25} - \frac{1}{25} ln (5)

= \frac{12}{25} - \frac{1}{25} ln (5)

\int_{1}^{2} \frac{ln ( x )}{x ^{6}} d x

\int_{0}^{\frac{π}{2}} (- 3 x^{2} + x - 2) d x

\int_{0}^{1} \frac{x ^{4}}{x ^{5} + 1} d x

\int_{- 2}^{0} (4 x^{3} + x^{2} - 2 x) d x

\int_{0}^{1} (2 t + 5) (e^{- t}) d t

What is the integral from 1 to 5 of (ln(x^2))/(x^3) ?
The integral from 1 to 5 of (ln(x^2))/(x^3) is 12/25-1/25 ln(5)