What is the integral of 12x^3(ln(4x))^2 ?

The integral of 12x^3(ln(4x))^2 is 12(1/4 x^4ln^2(4x)-1/2 (1/4 x^4ln(4x)-(x^4)/(16)))+C

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integral of 12x^3(ln(4x))^2

\int 12 x^{3} (ln (4 x))^{2} d x

12 (\frac{1}{4} x^{4} ln^{2} (4 x) - \frac{1}{2} (\frac{1}{4} x^{4} ln (4 x) - \frac{x ^{4}}{16})) + C

Solution steps

\int 12 x^{3} (ln (4 x))^{2} d x

Take the constant out:

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

= 12 \cdot \int x^{3} (ln (4 x))^{2} d x

Simplify

= 12 \cdot \int x^{3} ln^{2} (4 x) d x

Apply Integration By Parts

: \frac{1}{4} x^{4} ln^{2} (4 x) - \int \frac{1}{2} x^{3} ln (4 x) d x

= 12 (\frac{1}{4} x^{4} ln^{2} (4 x) - \int \frac{1}{2} x^{3} ln (4 x) d x)

\int \frac{1}{2} x^{3} ln (4 x) d x = \frac{1}{2} (\frac{1}{4} x^{4} ln (4 x) - \frac{x ^{4}}{16})

= 12 (\frac{1}{4} x^{4} ln^{2} (4 x) - \frac{1}{2} (\frac{1}{4} x^{4} ln (4 x) - \frac{x ^{4}}{16}))

Add a constant to the solution

= 12 (\frac{1}{4} x^{4} ln^{2} (4 x) - \frac{1}{2} (\frac{1}{4} x^{4} ln (4 x) - \frac{x ^{4}}{16})) + C

\int (3 x^{2} + 8) d x

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\frac{d}{d x} ((x^{2} + 10)^{6})

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

What is the integral of 12x^3(ln(4x))^2 ?
The integral of 12x^3(ln(4x))^2 is 12(1/4 x^4ln^2(4x)-1/2 (1/4 x^4ln(4x)-(x^4)/(16)))+C