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

The integral of (7x^8+2/3 x^4-x+4) is (7x^9)/9+(2x^5}{15}-\frac{x^2)/2+4x+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

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

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

\frac{7 x ^{9}}{9} + \frac{2 x ^{5}}{15} - \frac{x ^{2}}{2} + 4 x + C

Solution steps

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

\int 7 x^{8} d x = \frac{7 x ^{9}}{9}

\int \frac{2}{3} x^{4} d x = \frac{2 x ^{5}}{15}

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

\int 4 d x = 4 x

= \frac{7 x ^{9}}{9} + \frac{2 x ^{5}}{15} - \frac{x ^{2}}{2} + 4 x

Add a constant to the solution

= \frac{7 x ^{9}}{9} + \frac{2 x ^{5}}{15} - \frac{x ^{2}}{2} + 4 x + C

What is the integral of (7x^8+2/3 x^4-x+4) ?
The integral of (7x^8+2/3 x^4-x+4) is (7x^9)/9+(2x^5}{15}-\frac{x^2)/2+4x+C