What is the integral of 7x^6(x^7+2)^5 ?

The integral of 7x^6(x^7+2)^5 is (x^{42})/6+2x^{35}+10x^{28}+(80x^{21})/3+40x^{14}+32x^7+C

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integral of 7x^6(x^7+2)^5

\int 7 x^{6} (x^{7} + 2)^{5} d x

\frac{x ^{42}}{6} + 2 x^{35} + 10 x^{28} + \frac{80 x ^{21}}{3} + 40 x^{14} + 32 x^{7} + C

Solution steps

\int 7 x^{6} (x^{7} + 2)^{5} 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} + 2)^{5} d x

Expand

x^{6} (x^{7} + 2)^{5} : x^{41} + 10 x^{34} + 40 x^{27} + 80 x^{20} + 80 x^{13} + 32 x^{6}

= 7 \cdot \int x^{41} + 10 x^{34} + 40 x^{27} + 80 x^{20} + 80 x^{13} + 32 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^{41} d x + \int 10 x^{34} d x + \int 40 x^{27} d x + \int 80 x^{20} d x + \int 80 x^{13} d x + \int 32 x^{6} d x)

\int x^{41} d x = \frac{x ^{42}}{42}

\int 10 x^{34} d x = \frac{2 x ^{35}}{7}

\int 40 x^{27} d x = \frac{10 x ^{28}}{7}

\int 80 x^{20} d x = \frac{80 x ^{21}}{21}

\int 80 x^{13} d x = \frac{40 x ^{14}}{7}

\int 32 x^{6} d x = \frac{32 x ^{7}}{7}

= 7 (\frac{x ^{42}}{42} + \frac{2 x ^{35}}{7} + \frac{10 x ^{28}}{7} + \frac{80 x ^{21}}{21} + \frac{40 x ^{14}}{7} + \frac{32 x ^{7}}{7})

Simplify

7 (\frac{x ^{42}}{42} + \frac{2 x ^{35}}{7} + \frac{10 x ^{28}}{7} + \frac{80 x ^{21}}{21} + \frac{40 x ^{14}}{7} + \frac{32 x ^{7}}{7}) : \frac{x ^{42}}{6} + 2 x^{35} + 10 x^{28} + \frac{80 x ^{21}}{3} + 40 x^{14} + 32 x^{7}

= \frac{x ^{42}}{6} + 2 x^{35} + 10 x^{28} + \frac{80 x ^{21}}{3} + 40 x^{14} + 32 x^{7}

Add a constant to the solution

= \frac{x ^{42}}{6} + 2 x^{35} + 10 x^{28} + \frac{80 x ^{21}}{3} + 40 x^{14} + 32 x^{7} + C

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What is the integral of 7x^6(x^7+2)^5 ?
The integral of 7x^6(x^7+2)^5 is (x^{42})/6+2x^{35}+10x^{28}+(80x^{21})/3+40x^{14}+32x^7+C