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

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

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

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

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

Solution steps

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

Simplify

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

Apply u-substitution

: \int - \frac{2 a ln ^{2} ( u )}{u ^{2}} d u

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

Take the constant out:

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

= - 2 a \cdot \int \frac{ln ^{2} ( u )}{u ^{2}} d u

Apply Integration By Parts

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

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

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

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

Substitute back

u = \frac{2 a}{x}

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

Simplify

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

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

Add a constant to the solution

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

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