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

The integral of 3x(ln(x^3))^2 is 3(9/2 x^2ln^2(x)-3(3/2 x^2ln(x)-(3x^2)/4))+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

3 (\frac{9}{2} x^{2} ln^{2} (x) - 3 (\frac{3}{2} x^{2} ln (x) - \frac{3 x ^{2}}{4})) + C

Solution steps

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

Take the constant out:

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

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

Simplify

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

Apply Integration By Parts

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

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

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

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

\frac{1}{2} x^{2} ln^{2} (x^{3}) = \frac{9}{2} x^{2} ln^{2} (x)

= 3 (\frac{9}{2} x^{2} ln^{2} (x) - 3 (\frac{3}{2} x^{2} ln (x) - \frac{3 x ^{2}}{4}))

Add a constant to the solution

= 3 (\frac{9}{2} x^{2} ln^{2} (x) - 3 (\frac{3}{2} x^{2} ln (x) - \frac{3 x ^{2}}{4})) + C

\int x + 4 ln (x + 4) d x

e^{- \frac{x}{2}}

\int \frac{x ^{2}}{x ^{6} + 16} d x

\int cos (x) sin (x) + 2 d x

\int (a sin (n x) + b cos (m x)) d x

What is the integral of 3x(ln(x^3))^2 ?
The integral of 3x(ln(x^3))^2 is 3(9/2 x^2ln^2(x)-3(3/2 x^2ln(x)-(3x^2)/4))+C