What is the integral from 0 to 2 of (6-4ln(3-x))^2 ?

The integral from 0 to 2 of (6-4ln(3-x))^2 is 48ln^2(3)-240ln(3)+232

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integral from 0 to 2 of (6-4ln(3-x))^2

\int_{0}^{2} (6 - 4 ln (3 - x))^{2} d x

48 ln^{2} (3) - 240 ln (3) + 232

Decimal Notation

26.26660 \dots

Solution steps

\int_{0}^{2} (6 - 4 ln (3 - x))^{2} d x

Expand

(6 - 4 ln (3 - x))^{2} : 36 - 48 ln (3 - x) + 16 ln^{2} (3 - x)

= \int_{0}^{2} 36 - 48 ln (3 - x) + 16 ln^{2} (3 - 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_{0}^{2} 36 d x - \int_{0}^{2} 48 ln (3 - x) d x + \int_{0}^{2} 16 ln^{2} (3 - x) d x

\int_{0}^{2} 36 d x = 72

\int_{0}^{2} 48 ln (3 - x) d x = 48 (3 ln (3) - 2)

\int_{0}^{2} 16 ln^{2} (3 - x) d x = 16 (3 ln^{2} (3) - 6 ln (3) + 4)

= 72 - 48 (3 ln (3) - 2) + 16 (3 ln^{2} (3) - 6 ln (3) + 4)

Simplify

72 - 48 (3 ln (3) - 2) + 16 (3 ln^{2} (3) - 6 ln (3) + 4) : 48 ln^{2} (3) - 240 ln (3) + 232

= 48 ln^{2} (3) - 240 ln (3) + 232

\int 9 x d x

y y^{^{'}} = t

f (x) = ∣ x - 4 ∣

\frac{d}{d x} (\frac{1}{( 3 x ^{3} + 5 x + 4 ) ^{\frac{3}{4}}})

\frac{d}{d x} ((f (x) (6 x))^{2} + 3)

What is the integral from 0 to 2 of (6-4ln(3-x))^2 ?
The integral from 0 to 2 of (6-4ln(3-x))^2 is 48ln^2(3)-240ln(3)+232