What is the integral of ln(x)*1/x ?

The integral of ln(x)*1/x is (ln^2(x))/2+C

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integral of ln(x)*1/x

\int ln (x) \cdot \frac{1}{x} d x

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

Solution steps

\int ln (x) \frac{1}{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

\int (1 - x^{2} - x) x d x

\int \frac{x ^{4} + x - 16}{x ^{2} + 4} d x

ln (\frac{a ^{2} - z ^{2}}{a ^{2} + z ^{2}})

y = 3 (4 x^{2} + 4)

\frac{\partial}{\partial x} (z x + y)