What is the integral of x^{-1}ln(x) ?

The integral of x^{-1}ln(x) is (ln^2(x))/2+C

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integral of x^{-1}ln(x)

\int x^{- 1} ln (x) d x

\frac{ln ^{2} ( x )}{2} + C

Solution steps

\int x^{- 1} ln (x) d x

Simplify

x^{- 1} ln (x) : \frac{ln ( x )}{x}

= \int \frac{ln ( x )}{x} d x

Apply u-substitution

: \int u d u

= \int u d u

Apply the Power Rule

: \frac{u ^{2}}{2}

= \frac{u ^{2}}{2}

Substitute back

u = ln (x)

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

Add a constant to the solution

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

x \to - 3 lim (\frac{x + 3}{x + 3})

f (x) = 17 x

\int \frac{x ^{2} + 9}{x ^{2}} d x

\frac{8 t}{8 + t}

\int \frac{3 x ^{2}}{2} d x

What is the integral of x^{-1}ln(x) ?
The integral of x^{-1}ln(x) is (ln^2(x))/2+C