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

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

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integral of 1/x ln(x)

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

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

Solution steps

\int \frac{1}{x} ln (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 - (t - 6)^{2} d t

\frac{d}{d y} (x^{0.5} y^{0.5})

x^{\frac{7}{9}}

f (x) = x^{2} - 7, at x = 2

\frac{d x}{d t} = 0.9 x (1900 - x)