What is the integral of (x^5-7x)^2(5x^4-7) ?

The integral of (x^5-7x)^2(5x^4-7) is (x^{15})/3-7x^{11}+49x^7-(343x^3)/3+C

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integral of (x^5-7x)^2(5x^4-7)

\int (x^{5} - 7 x)^{2} (5 x^{4} - 7) d x

\frac{x ^{15}}{3} - 7 x^{11} + 49 x^{7} - \frac{343 x ^{3}}{3} + C

Solution steps

\int (x^{5} - 7 x)^{2} (5 x^{4} - 7) d x

Expand

(x^{5} - 7 x)^{2} (5 x^{4} - 7) : 5 x^{14} - 77 x^{10} + 343 x^{6} - 343 x^{2}

= \int 5 x^{14} - 77 x^{10} + 343 x^{6} - 343 x^{2} 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 5 x^{14} d x - \int 77 x^{10} d x + \int 343 x^{6} d x - \int 343 x^{2} d x

\int 5 x^{14} d x = \frac{x ^{15}}{3}

\int 77 x^{10} d x = 7 x^{11}

\int 343 x^{6} d x = 49 x^{7}

\int 343 x^{2} d x = \frac{343 x ^{3}}{3}

= \frac{x ^{15}}{3} - 7 x^{11} + 49 x^{7} - \frac{343 x ^{3}}{3}

Add a constant to the solution

= \frac{x ^{15}}{3} - 7 x^{11} + 49 x^{7} - \frac{343 x ^{3}}{3} + C

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What is the integral of (x^5-7x)^2(5x^4-7) ?
The integral of (x^5-7x)^2(5x^4-7) is (x^{15})/3-7x^{11}+49x^7-(343x^3)/3+C