What is the integral of (6x^8+9x^2-9x)/(3x^2) ?

The integral of (6x^8+9x^2-9x)/(3x^2) is 1/7 (2x^7+21x-21ln|x|)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (6x^8+9x^2-9x)/(3x^2)

\int \frac{6 x ^{8} + 9 x ^{2} - 9 x}{3 x ^{2}} d x

\frac{1}{7} (2 x^{7} + 21 x - 21 ln ∣ x ∣) + C

Solution steps

\int \frac{6 x ^{8} + 9 x ^{2} - 9 x}{3 x ^{2}} d x

Take the constant out:

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

= \frac{1}{3} \cdot \int \frac{6 x ^{8} + 9 x ^{2} - 9 x}{x ^{2}} d x

Expand

\frac{6 x ^{8} + 9 x ^{2} - 9 x}{x ^{2}} : 6 x^{6} + 9 - \frac{9}{x}

= \frac{1}{3} \cdot \int 6 x^{6} + 9 - \frac{9}{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}{3} (\int 6 x^{6} d x + \int 9 d x - \int \frac{9}{x} d x)

\int 6 x^{6} d x = \frac{6 x ^{7}}{7}

\int 9 d x = 9 x

\int \frac{9}{x} d x = 9 ln ∣ x ∣

= \frac{1}{3} (\frac{6 x ^{7}}{7} + 9 x - 9 ln ∣ x ∣)

Simplify

\frac{1}{3} (\frac{6 x ^{7}}{7} + 9 x - 9 ln ∣ x ∣) : \frac{1}{7} (2 x^{7} + 21 x - 21 ln ∣ x ∣)

= \frac{1}{7} (2 x^{7} + 21 x - 21 ln ∣ x ∣)

Add a constant to the solution

= \frac{1}{7} (2 x^{7} + 21 x - 21 ln ∣ x ∣) + C

\int \frac{x + 1}{( x ^{2} + 1 )} d x

\int 4 x^{3} + sin (x) d x

\int \frac{3 x ^{3} - x ^{2} + 6 x + 24}{x ^{2} - 2 x + 5} d x

\int (\frac{( 2 + x ) ^{2}}{x}) d x

\int \frac{e ^{4 x}}{( 1 + 3 e ^{4 x} ) ^{2}} d x

What is the integral of (6x^8+9x^2-9x)/(3x^2) ?
The integral of (6x^8+9x^2-9x)/(3x^2) is 1/7 (2x^7+21x-21ln|x|)+C