What is the integral from 0 to 1 of x^{-3}ln(x) ?

The integral from 0 to 1 of x^{-3}ln(x) is diverges

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integral from 0 to 1 of x^{-3}ln(x)

\int_{0}^{1} x^{- 3} ln (x) d x

diverges

Solution steps

\int_{0}^{1} x^{- 3} ln (x) d x

Simplify

x^{- 3} ln (x) : \frac{ln ( x )}{x ^{3}}

= \int_{0}^{1} \frac{ln ( x )}{x ^{3}} d x

Apply Integration By Parts

: [- \frac{ln ( x )}{2 x ^{2}} - \int - \frac{1}{2 x ^{3}} d x]_{0}^{1}

= [- \frac{ln ( x )}{2 x ^{2}} - \int - \frac{1}{2 x ^{3}} d x]_{0}^{1}

\int - \frac{1}{2 x ^{3}} d x = \frac{1}{4 x ^{2}}

= [- \frac{ln ( x )}{2 x ^{2}} - \frac{1}{4 x ^{2}}]_{0}^{1}

Compute the boundaries:

diverges

= diverges

\int_{- 3}^{3} - x^{2} + 10 d x

\int_{b}^{\infty} \frac{1}{1 + x ^{2}} d x

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

\int_{0}^{t} sin (x) f (t - x) d x

\int_{0}^{0.9} \frac{1}{( 1 - x ) ^{2}} d x

What is the integral from 0 to 1 of x^{-3}ln(x) ?
The integral from 0 to 1 of x^{-3}ln(x) is diverges