What is the integral of 8x(x^7-1/8) ?

The integral of 8x(x^7-1/8) is 8((x^9)/9-(x^2)/(16))+C

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integral of 8x(x^7-1/8)

\int 8 x (x^{7} - \frac{1}{8}) d x

8 (\frac{x ^{9}}{9} - \frac{x ^{2}}{16}) + C

Solution steps

\int 8 x (x^{7} - \frac{1}{8}) d x

Take the constant out:

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

= 8 \cdot \int x (x^{7} - \frac{1}{8}) d x

Expand

x (x^{7} - \frac{1}{8}) : x^{8} - \frac{x}{8}

= 8 \cdot \int x^{8} - \frac{x}{8} 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

= 8 (\int x^{8} d x - \int \frac{x}{8} d x)

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

\int \frac{x}{8} d x = \frac{x ^{2}}{16}

= 8 (\frac{x ^{9}}{9} - \frac{x ^{2}}{16})

Add a constant to the solution

= 8 (\frac{x ^{9}}{9} - \frac{x ^{2}}{16}) + C

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What is the integral of 8x(x^7-1/8) ?
The integral of 8x(x^7-1/8) is 8((x^9)/9-(x^2)/(16))+C