What is the integral of (x^3-1)^5 ?

The integral of (x^3-1)^5 is (x^{16}}{16}-(5x^{13})/(13)+x^{10}-\frac{10x^7)/7+(5x^4)/4-x+C

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integral of (x^3-1)^5

integral

(x^{3} - 1)^{5}

\frac{x ^{16}}{16} - \frac{5 x ^{13}}{13} + x^{10} - \frac{10 x ^{7}}{7} + \frac{5 x ^{4}}{4} - x + C

Solution steps

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

Expand

(x^{3} - 1)^{5} : x^{15} - 5 x^{12} + 10 x^{9} - 10 x^{6} + 5 x^{3} - 1

= \int x^{15} - 5 x^{12} + 10 x^{9} - 10 x^{6} + 5 x^{3} - 1 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 x^{15} d x - \int 5 x^{12} d x + \int 10 x^{9} d x - \int 10 x^{6} d x + \int 5 x^{3} d x - \int 1 d x

\int x^{15} d x = \frac{x ^{16}}{16}

\int 5 x^{12} d x = \frac{5 x ^{13}}{13}

\int 10 x^{9} d x = x^{10}

\int 10 x^{6} d x = \frac{10 x ^{7}}{7}

\int 5 x^{3} d x = \frac{5 x ^{4}}{4}

\int 1 d x = x

= \frac{x ^{16}}{16} - \frac{5 x ^{13}}{13} + x^{10} - \frac{10 x ^{7}}{7} + \frac{5 x ^{4}}{4} - x

Add a constant to the solution

= \frac{x ^{16}}{16} - \frac{5 x ^{13}}{13} + x^{10} - \frac{10 x ^{7}}{7} + \frac{5 x ^{4}}{4} - x + C

in t e g r a l \frac{x ^{2}}{x + 4}

in t e g r a l \frac{2 ln ( x )}{x}

in t e g r a l \frac{1}{( y ^{2} + 6 y + 5 )}

in t e g r a l x^{2} + 3 x

in t e g r a l e^{x} (1 + x)

What is the integral of (x^3-1)^5 ?
The integral of (x^3-1)^5 is (x^{16}}{16}-(5x^{13})/(13)+x^{10}-\frac{10x^7)/7+(5x^4)/4-x+C