What is the integral of 16xln(5x) ?

The integral of 16xln(5x) is 16(1/2 x^2ln(5x)-(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 16xln(5x)

\int 16 x ln (5 x) d x

16 (\frac{1}{2} x^{2} ln (5 x) - \frac{x ^{2}}{4}) + C

Solution steps

\int 16 x ln (5 x) d x

Take the constant out:

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

= 16 \cdot \int x ln (5 x) d x

Apply Integration By Parts

: \frac{1}{2} x^{2} ln (5 x) - \int \frac{x}{2} d x

= 16 (\frac{1}{2} x^{2} ln (5 x) - \int \frac{x}{2} d x)

\int \frac{x}{2} d x = \frac{x ^{2}}{4}

= 16 (\frac{1}{2} x^{2} ln (5 x) - \frac{x ^{2}}{4})

Add a constant to the solution

= 16 (\frac{1}{2} x^{2} ln (5 x) - \frac{x ^{2}}{4}) + C

\frac{\partial}{\partial x} (x (x^{2} + y^{2})^{- 1})

f (x) = 4 x cos (x)

x - 9

\int \frac{1}{2 x ^{2} + 4 x + 7} d x

\frac{1}{6} y^{^{''}} + 6 y = 4 tan (6 t) - \frac{1}{2} e^{3 t}

What is the integral of 16xln(5x) ?
The integral of 16xln(5x) is 16(1/2 x^2ln(5x)-(x^2)/4)+C