What is the integral of ((7x^6-8x^4+9x^3)/(2x^2)) ?

The integral of ((7x^6-8x^4+9x^3)/(2x^2)) is 1/2 ((7x^5)/5-(8x^3)/3+(9x^2)/2)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of ((7x^6-8x^4+9x^3)/(2x^2))

\int (\frac{7 x ^{6} - 8 x ^{4} + 9 x ^{3}}{2 x ^{2}}) d x

\frac{1}{2} (\frac{7 x ^{5}}{5} - \frac{8 x ^{3}}{3} + \frac{9 x ^{2}}{2}) + C

Solution steps

\int \frac{7 x ^{6} - 8 x ^{4} + 9 x ^{3}}{2 x ^{2}} d x

Take the constant out:

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

= \frac{1}{2} \cdot \int \frac{7 x ^{6} - 8 x ^{4} + 9 x ^{3}}{x ^{2}} d x

Expand

\frac{7 x ^{6} - 8 x ^{4} + 9 x ^{3}}{x ^{2}} : \frac{7 x ^{6}}{x ^{2}} - \frac{8 x ^{4}}{x ^{2}} + \frac{9 x ^{3}}{x ^{2}}

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}{2} (\int \frac{7 x ^{6}}{x ^{2}} d x - \int \frac{8 x ^{4}}{x ^{2}} d x + \int \frac{9 x ^{3}}{x ^{2}} d x)

\int \frac{7 x ^{6}}{x ^{2}} d x = \frac{7 x ^{5}}{5}

\int \frac{8 x ^{4}}{x ^{2}} d x = \frac{8 x ^{3}}{3}

\int \frac{9 x ^{3}}{x ^{2}} d x = \frac{9 x ^{2}}{2}

= \frac{1}{2} (\frac{7 x ^{5}}{5} - \frac{8 x ^{3}}{3} + \frac{9 x ^{2}}{2})

Add a constant to the solution

= \frac{1}{2} (\frac{7 x ^{5}}{5} - \frac{8 x ^{3}}{3} + \frac{9 x ^{2}}{2}) + C

\int \frac{x}{4 + 4 x - x ^{2}} d x

\int \frac{x ^{2}}{9 + x} d x

\int \frac{x + 1}{3 x ^{2} + 6 x - 5} d x

\int 2 x (x^{2} - 4)^{6} d x

\int \frac{x}{x ^{2} + 10 x + 50} d x

What is the integral of ((7x^6-8x^4+9x^3)/(2x^2)) ?
The integral of ((7x^6-8x^4+9x^3)/(2x^2)) is 1/2 ((7x^5)/5-(8x^3)/3+(9x^2)/2)+C