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

The integral of (ln^2(x)+ln(x))/(x^2) is -(ln^2(x))/x-(2ln(x))/x-3/x-(ln(x))/x+C

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

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

- \frac{ln ^{2} ( x )}{x} - \frac{2 ln ( x )}{x} - \frac{3}{x} - \frac{ln ( x )}{x} + C

Solution steps

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

Expand

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

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

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

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

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

Simplify

- \frac{ln ^{2} ( x )}{x} + 2 (- \frac{ln ( x )}{x} - \frac{1}{x}) - \frac{ln ( x )}{x} - \frac{1}{x} : - \frac{ln ^{2} ( x )}{x} - \frac{2 ln ( x )}{x} - \frac{3}{x} - \frac{ln ( x )}{x}

= - \frac{ln ^{2} ( x )}{x} - \frac{2 ln ( x )}{x} - \frac{3}{x} - \frac{ln ( x )}{x}

Add a constant to the solution

= - \frac{ln ^{2} ( x )}{x} - \frac{2 ln ( x )}{x} - \frac{3}{x} - \frac{ln ( x )}{x} + C

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What is the integral of (ln^2(x)+ln(x))/(x^2) ?
The integral of (ln^2(x)+ln(x))/(x^2) is -(ln^2(x))/x-(2ln(x))/x-3/x-(ln(x))/x+C