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

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

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

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

x (ln (x) - 1) + C

Solution steps

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

Take the constant out:

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

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

Apply Integration By Parts

: x ln (x^{2}) - \int 2 d x

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

\int 2 d x = 2 x

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

Simplify

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

= x (ln (x) - 1)

Add a constant to the solution

= x (ln (x) - 1) + C

x^{3} + 8 x - 1, - 1

\frac{\partial}{\partial y} (29 - x^{2} - 4 y^{2})

\frac{d}{d x} (4^{x} lo g_{4} (x))

F (x) = e^{x^{2}}

\frac{d}{d x} (- x^{2})

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