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

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

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

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

\frac{1}{2} ln (x) + C

Solution steps

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

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

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

Apply u-substitution

: \int \frac{1 - 2 e ^{u}}{2} + e^{u} d u

= \int \frac{1 - 2 e ^{u}}{2} + e^{u} d u

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int \frac{1 - 2 e ^{u}}{2} d u + \int e^{u} d u

\int \frac{1 - 2 e ^{u}}{2} d u = \frac{1}{2} u - e^{u}

\int e^{u} d u = e^{u}

= \frac{1}{2} u - e^{u} + e^{u}

Substitute back

u = ln (x)

= \frac{1}{2} ln (x) - e^{l n (x)} + e^{l n (x)}

Simplify

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

Add a constant to the solution

= \frac{1}{2} ln (x) + C

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