What is the integral of 10x(ln(x))^2 ?

The integral of 10x(ln(x))^2 is 10(1/2 x^2ln^2(x)-1/2 x^2ln(x)+(x^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 10x(ln(x))^2

\int 10 x (ln (x))^{2} d x

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

Solution steps

\int 10 x (ln (x))^{2} d x

Take the constant out:

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

= 10 \cdot \int x (ln (x))^{2} d x

Apply Integration By Parts

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

= 10 (\frac{1}{2} x^{2} ln^{2} (x) - \int x ln (x) d x)

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

= 10 (\frac{1}{2} x^{2} ln^{2} (x) - (\frac{1}{2} x^{2} ln (x) - \frac{x ^{2}}{4}))

Simplify

= 10 (\frac{1}{2} x^{2} ln^{2} (x) - \frac{1}{2} x^{2} ln (x) + \frac{x ^{2}}{4})

Add a constant to the solution

= 10 (\frac{1}{2} x^{2} ln^{2} (x) - \frac{1}{2} x^{2} ln (x) + \frac{x ^{2}}{4}) + C

\int (5 x^{2} + 6)^{2} d x

\int (x^{3} + 1)^{10} d x

\int \frac{( a + b x ^{8} )}{9 a x + b x ^{9}} d x

\int (8 sec (x) tan (x) - 9 sec^{2} (x)) d x

\int \frac{1}{x ^{1.2}} d x

What is the integral of 10x(ln(x))^2 ?
The integral of 10x(ln(x))^2 is 10(1/2 x^2ln^2(x)-1/2 x^2ln(x)+(x^2)/4)+C