What is the integral of 5xlog_{10}(3-x) ?

The integral of 5xlog_{10}(3-x) is 5(1/2 x^2log_{10}(3-x)+1/(2ln(10))(-(x^2)/2-3x-9ln|-x+3|))+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of 5xlog_{10}(3-x)

\int 5 x lo g_{10} (3 - x) d x

5 (\frac{1}{2} x^{2} lo g_{10} (3 - x) + \frac{1}{2 ln ( 10 )} (- \frac{x ^{2}}{2} - 3 x - 9 ln ∣ - x + 3 ∣)) + C

Solution steps

\int 5 x lo g_{10} (3 - x) d x

Take the constant out:

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

= 5 \cdot \int x lo g_{10} (3 - x) d x

Apply Integration By Parts

: \frac{1}{2} x^{2} lo g_{10} (3 - x) - \int - \frac{x ^{2}}{2 ln ( 10 ) ( 3 - x )} d x

= 5 (\frac{1}{2} x^{2} lo g_{10} (3 - x) - \int - \frac{x ^{2}}{2 ln ( 10 ) ( 3 - x )} d x)

\int - \frac{x ^{2}}{2 ln ( 10 ) ( 3 - x )} d x = - \frac{1}{2 ln ( 10 )} (- \frac{x ^{2}}{2} - 3 x - 9 ln ∣ - x + 3 ∣)

= 5 (\frac{1}{2} x^{2} lo g_{10} (3 - x) - (- \frac{1}{2 ln ( 10 )} (- \frac{x ^{2}}{2} - 3 x - 9 ln ∣ - x + 3 ∣)))

Simplify

= 5 (\frac{1}{2} x^{2} lo g_{10} (3 - x) + \frac{1}{2 ln ( 10 )} (- \frac{x ^{2}}{2} - 3 x - 9 ln ∣ - x + 3 ∣))

Add a constant to the solution

= 5 (\frac{1}{2} x^{2} lo g_{10} (3 - x) + \frac{1}{2 ln ( 10 )} (- \frac{x ^{2}}{2} - 3 x - 9 ln ∣ - x + 3 ∣)) + C

x, 0, 1

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

\frac{d}{d x} (5 x^{3})

\frac{d}{d x} (\frac{4}{3 x ^{3}})

\int x^{3} x^{2} - 5 d x

What is the integral of 5xlog_{10}(3-x) ?
The integral of 5xlog_{10}(3-x) is 5(1/2 x^2log_{10}(3-x)+1/(2ln(10))(-(x^2)/2-3x-9ln|-x+3|))+C