What is the integral of 7x^6(x^7-5)^3 ?

The integral of 7x^6(x^7-5)^3 is (x^{28})/4-5x^{21}+(75x^{14})/2-125x^7+C

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integral of 7x^6(x^7-5)^3

\int 7 x^{6} (x^{7} - 5)^{3} d x

\frac{x ^{28}}{4} - 5 x^{21} + \frac{75 x ^{14}}{2} - 125 x^{7} + C

Solution steps

\int 7 x^{6} (x^{7} - 5)^{3} d x

Take the constant out:

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

= 7 \cdot \int x^{6} (x^{7} - 5)^{3} d x

Expand

x^{6} (x^{7} - 5)^{3} : x^{27} - 15 x^{20} + 75 x^{13} - 125 x^{6}

= 7 \cdot \int x^{27} - 15 x^{20} + 75 x^{13} - 125 x^{6} 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

= 7 (\int x^{27} d x - \int 15 x^{20} d x + \int 75 x^{13} d x - \int 125 x^{6} d x)

\int x^{27} d x = \frac{x ^{28}}{28}

\int 15 x^{20} d x = \frac{5 x ^{21}}{7}

\int 75 x^{13} d x = \frac{75 x ^{14}}{14}

\int 125 x^{6} d x = \frac{125 x ^{7}}{7}

= 7 (\frac{x ^{28}}{28} - \frac{5 x ^{21}}{7} + \frac{75 x ^{14}}{14} - \frac{125 x ^{7}}{7})

Simplify

7 (\frac{x ^{28}}{28} - \frac{5 x ^{21}}{7} + \frac{75 x ^{14}}{14} - \frac{125 x ^{7}}{7}) : \frac{x ^{28}}{4} - 5 x^{21} + \frac{75 x ^{14}}{2} - 125 x^{7}

= \frac{x ^{28}}{4} - 5 x^{21} + \frac{75 x ^{14}}{2} - 125 x^{7}

Add a constant to the solution

= \frac{x ^{28}}{4} - 5 x^{21} + \frac{75 x ^{14}}{2} - 125 x^{7} + C

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What is the integral of 7x^6(x^7-5)^3 ?
The integral of 7x^6(x^7-5)^3 is (x^{28})/4-5x^{21}+(75x^{14})/2-125x^7+C