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

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

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

integral

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

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

Solution steps

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

Apply Integration By Parts

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

\int - \frac{2}{x ^{2}} d x = \frac{2}{x}

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

Apply log rule

lo g_{a} (x^{b}) = b \cdot lo g_{a} (x),

assuming

x \geq 0

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

Add a constant to the solution

= - \frac{2 ln ( x )}{x} - \frac{2}{x} + C

in t e g r a l \frac{tan ( x )}{sin ^{2} ( x )}

in t e g r a l \frac{x ^{2}}{4 a + 2 a}

in t e g r a l \frac{( x - 1 )}{( x ^{2} + 1 )}

in t e g r a l \frac{1}{( x ^{2} - 4 )}

in t e g r a l \frac{4 x ^{4}}{( 1 + x ^{2} )}

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