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

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

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integral of (ln(x^2))/x

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

ln^{2} (x) + C

Solution steps

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

Apply u-substitution

: \int \frac{ln ( u )}{2 u} d u

= \int \frac{ln ( u )}{2 u} d u

Take the constant out:

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

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

Apply u-substitution

: \int v d v

= \frac{1}{2} \cdot \int v d v

Apply the Power Rule

: \frac{v ^{2}}{2}

= \frac{1}{2} \cdot \frac{v ^{2}}{2}

Substitute back

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

Simplify

\frac{1}{2} \cdot \frac{ln ^{2} ( x ^{2} )}{2} : ln^{2} (x)

= ln^{2} (x)

Add a constant to the solution

= ln^{2} (x) + C

\int_{0}^{1} (arctan (2 x)) d x

f (x) = 7 x^{2} tan (x)

x y^{^{'}} + y = 7 x y^{2}

\int 6 x 3 x^{2} + 2 d x

f (x) = x + 7