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

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

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integral of (1+ln(x^2))/x

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

ln ∣ x ∣ + ln^{2} (x) + C

Solution steps

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

Expand

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

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

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

\int \frac{1}{x} d x = ln ∣ x ∣

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

= ln ∣ x ∣ + ln^{2} (x)

Add a constant to the solution

= ln ∣ x ∣ + ln^{2} (x) + C

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