What is the integral from 3 to 9 of t^2ln(4t) ?

The integral from 3 to 9 of t^2ln(4t) is 486ln(6)-78-9ln(12)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral from 3 to 9 of t^2ln(4t)

\int_{3}^{9} t^{2} ln (4 t) d t

486 ln (6) - 78 - 9 ln (12)

Decimal Notation

770.43094 \dots

Solution steps

\int_{3}^{9} t^{2} ln (4 t) d t

Apply Integration By Parts

: [\frac{1}{3} t^{3} ln (4 t) - \int \frac{t ^{2}}{3} d t]_{3}^{9}

= [\frac{1}{3} t^{3} ln (4 t) - \int \frac{t ^{2}}{3} d t]_{3}^{9}

\int \frac{t ^{2}}{3} d t = \frac{t ^{3}}{9}

= [\frac{1}{3} t^{3} ln (4 t) - \frac{t ^{3}}{9}]_{3}^{9}

Compute the boundaries:

486 ln (6) - 78 - 9 ln (12)

= 486 ln (6) - 78 - 9 ln (12)

\frac{d}{d x} (\frac{( sin ( x ) )}{1 + cos ( x )})

x \to 0 + lim (x ln (3 x))

\frac{6 x}{5 - cot ( x )}

e^{2 x^{2 x}}

x \to 1 lim ((x - 1)^{- 1})

What is the integral from 3 to 9 of t^2ln(4t) ?
The integral from 3 to 9 of t^2ln(4t) is 486ln(6)-78-9ln(12)