What is the integral of 3x^7+4x^5-2x^3-x+1 ?

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of 3x^7+4x^5-2x^3-x+1

\int 3 x^{7} + 4 x^{5} - 2 x^{3} - x + 1 d x

\frac{3 x ^{8}}{8} + \frac{2 x ^{6}}{3} - \frac{x ^{4}}{2} - \frac{x ^{2}}{2} + x + C

Solution steps

\int 3 x^{7} + 4 x^{5} - 2 x^{3} - x + 1 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

= \int 3 x^{7} d x + \int 4 x^{5} d x - \int 2 x^{3} d x - \int x d x + \int 1 d x

\int 3 x^{7} d x = \frac{3 x ^{8}}{8}

\int 4 x^{5} d x = \frac{2 x ^{6}}{3}

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

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

\int 1 d x = x

= \frac{3 x ^{8}}{8} + \frac{2 x ^{6}}{3} - \frac{x ^{4}}{2} - \frac{x ^{2}}{2} + x

Add a constant to the solution

= \frac{3 x ^{8}}{8} + \frac{2 x ^{6}}{3} - \frac{x ^{4}}{2} - \frac{x ^{2}}{2} + x + C

\frac{d}{d x} (\frac{( 2 x - 1 ) ^{3}}{( x - 3 ) ^{2}})

\frac{\partial}{\partial x} (z e^{x y})

\frac{d}{d x} (\frac{1400}{( 1 + 6 e ^{- 0.02 x} )})

f (x) = x^{- 2}

\frac{d}{d x} (\frac{sin ( x )}{x y})

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