What is the integral of ((1-x^8))/((1-x)) ?

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

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integral of ((1-x^8))/((1-x))

integral

\frac{( 1 - x ^{8} )}{( 1 - x )}

\frac{x ^{8}}{8} + \frac{x ^{7}}{7} + \frac{x ^{6}}{6} + \frac{x ^{5}}{5} + \frac{x ^{4}}{4} + \frac{x ^{3}}{3} + \frac{x ^{2}}{2} + x + C

Solution steps

\int \frac{1 - x ^{8}}{1 - x} d x

Long division

\frac{1 - x ^{8}}{1 - x} : x^{7} + x^{6} + x^{5} + x^{4} + x^{3} + x^{2} + x + 1

= \int x^{7} + x^{6} + x^{5} + x^{4} + x^{3} + x^{2} + 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 x^{7} d x + \int x^{6} d x + \int x^{5} d x + \int x^{4} d x + \int x^{3} d x + \int x^{2} d x + \int x d x + \int 1 d x

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

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

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

\int x^{4} d x = \frac{x ^{5}}{5}

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

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

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

\int 1 d x = x

= \frac{x ^{8}}{8} + \frac{x ^{7}}{7} + \frac{x ^{6}}{6} + \frac{x ^{5}}{5} + \frac{x ^{4}}{4} + \frac{x ^{3}}{3} + \frac{x ^{2}}{2} + x

Add a constant to the solution

= \frac{x ^{8}}{8} + \frac{x ^{7}}{7} + \frac{x ^{6}}{6} + \frac{x ^{5}}{5} + \frac{x ^{4}}{4} + \frac{x ^{3}}{3} + \frac{x ^{2}}{2} + x + C

f (8 - 2 x)

\frac{1}{1 + e ^{x}}

\frac{e ^{x}}{( e ^{x} + 2 )}

2 x (4 - x^{2})

\frac{1}{( x ^{- 0} 0.5 + x ^{0.5} )}

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