What is the integral of (x^2-8)/(8x) ?

The integral of (x^2-8)/(8x) is (x^2)/(16)-ln|x|+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (x^2-8)/(8x)

\int \frac{x ^{2} - 8}{8 x} d x

\frac{x ^{2}}{16} - ln ∣ x ∣ + C

Solution steps

\int \frac{x ^{2} - 8}{8 x} d x

Take the constant out:

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

= \frac{1}{8} \cdot \int \frac{x ^{2} - 8}{x} d x

Expand

\frac{x ^{2} - 8}{x} : x - \frac{8}{x}

= \frac{1}{8} \cdot \int x - \frac{8}{x} d x

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \frac{1}{8} (\int x d x - \int \frac{8}{x} d x)

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

\int \frac{8}{x} d x = 8 ln ∣ x ∣

= \frac{1}{8} (\frac{x ^{2}}{2} - 8 ln ∣ x ∣)

Simplify

\frac{1}{8} (\frac{x ^{2}}{2} - 8 ln ∣ x ∣) : \frac{x ^{2}}{16} - ln ∣ x ∣

= \frac{x ^{2}}{16} - ln ∣ x ∣

Add a constant to the solution

= \frac{x ^{2}}{16} - ln ∣ x ∣ + C

\frac{d}{d x} (ln (arctan (9 x^{4})))

f (x) = ln (x)

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

k \to 0 lim (\frac{( 2 z + 3 k ) ^{3} - 4 k ^{2} z}{2 z ( 2 z - k ) ^{2}})

\int_{2}^{\infty} \frac{2}{x ^{2} - x} d x

What is the integral of (x^2-8)/(8x) ?
The integral of (x^2-8)/(8x) is (x^2)/(16)-ln|x|+C