What is the integral of (x^3-7x)^4(3x^2-7) ?

The integral of (x^3-7x)^4(3x^2-7) is (x^{15})/5-7x^{13}+98x^{11}-686x^9+2401x^7-(16807x^5)/5+C

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integral of (x^3-7x)^4(3x^2-7)

\int (x^{3} - 7 x)^{4} (3 x^{2} - 7) d x

\frac{x ^{15}}{5} - 7 x^{13} + 98 x^{11} - 686 x^{9} + 2401 x^{7} - \frac{16807 x ^{5}}{5} + C

Solution steps

\int (x^{3} - 7 x)^{4} (3 x^{2} - 7) d x

Expand

(x^{3} - 7 x)^{4} (3 x^{2} - 7) : 3 x^{14} - 91 x^{12} + 1078 x^{10} - 6174 x^{8} + 16807 x^{6} - 16807 x^{4}

= \int 3 x^{14} - 91 x^{12} + 1078 x^{10} - 6174 x^{8} + 16807 x^{6} - 16807 x^{4} 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 3 x^{14} d x - \int 91 x^{12} d x + \int 1078 x^{10} d x - \int 6174 x^{8} d x + \int 16807 x^{6} d x - \int 16807 x^{4} d x

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

\int 91 x^{12} d x = 7 x^{13}

\int 1078 x^{10} d x = 98 x^{11}

\int 6174 x^{8} d x = 686 x^{9}

\int 16807 x^{6} d x = 2401 x^{7}

\int 16807 x^{4} d x = \frac{16807 x ^{5}}{5}

= \frac{x ^{15}}{5} - 7 x^{13} + 98 x^{11} - 686 x^{9} + 2401 x^{7} - \frac{16807 x ^{5}}{5}

Add a constant to the solution

= \frac{x ^{15}}{5} - 7 x^{13} + 98 x^{11} - 686 x^{9} + 2401 x^{7} - \frac{16807 x ^{5}}{5} + C

cos (3 x) 2.5

\frac{\partial}{\partial x} (6 x + y^{2})

n = 1 \sum \infty \frac{6 n !}{n ^{n}}

\frac{d}{d x} (\frac{x ^{2} - 5}{1 - 3 x ^{2}})

6 (sin (x^{2}) + 2 x^{2} cos (x^{2}))

What is the integral of (x^3-7x)^4(3x^2-7) ?
The integral of (x^3-7x)^4(3x^2-7) is (x^{15})/5-7x^{13}+98x^{11}-686x^9+2401x^7-(16807x^5)/5+C