What is the integral of 2ln(x)*1/x ?

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

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integral of 2ln(x)*1/x

\int 2 ln (x) \cdot \frac{1}{x} d x

ln^{2} (x) + C

Solution steps

\int 2 ln (x) \frac{1}{x} d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 2 \cdot \int ln (x) \frac{1}{x} d x

Apply u-substitution

: \int u d u

= 2 \cdot \int u d u

Apply the Power Rule

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

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

Substitute back

u = ln (x)

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

Simplify

2 \cdot \frac{ln ^{2} ( x )}{2} : ln^{2} (x)

= ln^{2} (x)

Add a constant to the solution

= ln^{2} (x) + C

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