What is the integral of (ln(4x))^2 ?

The integral of (ln(4x))^2 is xln^2(4x)-2(xln(4x)-x)+C

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integral of (ln(4x))^2

\int (ln (4 x))^{2} d x

x ln^{2} (4 x) - 2 (x ln (4 x) - x) + C

Solution steps

\int (ln (4 x))^{2} d x

Apply Integration By Parts

: x ln^{2} (4 x) - \int 2 ln (4 x) d x

= x ln^{2} (4 x) - \int 2 ln (4 x) d x

\int 2 ln (4 x) d x = 2 (x ln (4 x) - x)

= x ln^{2} (4 x) - 2 (x ln (4 x) - x)

Add a constant to the solution

= x ln^{2} (4 x) - 2 (x ln (4 x) - x) + C

\int_{2}^{4} 1 - (x - 3)^{2} d x

\frac{\partial}{\partial x} (6 q (r, t, x)^{2} + 7 r^{3} - \frac{8}{t} - q (r, t, x) t)

y y^{^{'}} + x = 0

x^{\frac{3}{4}}

\int y^{2} (1 - 2 x) d y

What is the integral of (ln(4x))^2 ?
The integral of (ln(4x))^2 is xln^2(4x)-2(xln(4x)-x)+C