How do you evaluate the indefinite integral $\int (6 x^{7}) d x$ $int (6x^7)dx$ ?

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How do you evaluate the indefinite integral $\int (6 x^{7}) d x$ $int (6x^7)dx$ ?

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Answer 1 · 2018-05-18T09:19:58

$\int (6 x^{7}) d x = \frac{3 x^{8}}{4} + C$ $int(6x^7)dx = (3x^8)/4 + C$

Explanation:

For this indefinite integral we can apply the power rule.

The power rule states: $\int x^{n} d x = \frac{x^{n + 1}}{n + 1} + C$ $intx^ndx = (x^(n+1))/(n+1) + C$

So, when we plug in our values we get: $\int (6 x^{7}) d x = \frac{6 x^{7 + 1}}{7 + 1}$ $int(6x^7)dx = (6x^(7+1))/(7+1)$

$= \frac{6 x^{8}}{8}$ $= (6x^8)/(8)$

$= \frac{3 x^{8}}{4}$ $= (3x^8)/4$

$= \frac{3 x^{8}}{4} + C$ $= (3x^8)/4 + C$

Answer 2 · 2018-05-18T10:11:59

$\frac{3}{4} x^{8} + c$ $3/4x^8+c$

Explanation:

$integrate using the power rule$ $"integrate using the "color(blue)"power rule"$

$∙ x \int (a x^{n}) = \frac{a}{n + 1} x^{n + 1} x; n \neq - 1$ $•color(white)(x)int(ax^n)=a/(n+1)x^(n+1)color(white)(x);n!=-1$

$\Rightarrow \int (6 x^{7}) d x$ $rArrint(6x^7)dx$

$= \frac{6}{8} x^{(7 + 1)} + c = \frac{3}{4} x^{8} + c$ $=6/8x^((7+1))+c=3/4x^8+c$

$where c is the constant of integration$ $"where c is the constant of integration"$

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How do you evaluate the indefinite integral $\int (6 x^{7}) d x$ $int (6x^7)dx$ ?

2 Answers

Explanation:

Explanation:

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How do you evaluate the indefinite integral $\int (6 x^{7}) d x$ $int (6x^7)dx$ ?