What is the integral of x^3(x^4-1)^2 ?

The integral of x^3(x^4-1)^2 is (x^{12}}{12}-\frac{x^8)/4+(x^4)/4+C

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integral of x^3(x^4-1)^2

\int x^{3} (x^{4} - 1)^{2} d x

\frac{x ^{12}}{12} - \frac{x ^{8}}{4} + \frac{x ^{4}}{4} + C

Solution steps

\int x^{3} (x^{4} - 1)^{2} d x

Expand

x^{3} (x^{4} - 1)^{2} : x^{11} - 2 x^{7} + x^{3}

= \int x^{11} - 2 x^{7} + x^{3} 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^{11} d x - \int 2 x^{7} d x + \int x^{3} d x

\int x^{11} d x = \frac{x ^{12}}{12}

\int 2 x^{7} d x = \frac{x ^{8}}{4}

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

= \frac{x ^{12}}{12} - \frac{x ^{8}}{4} + \frac{x ^{4}}{4}

Add a constant to the solution

= \frac{x ^{12}}{12} - \frac{x ^{8}}{4} + \frac{x ^{4}}{4} + C

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What is the integral of x^3(x^4-1)^2 ?
The integral of x^3(x^4-1)^2 is (x^{12}}{12}-\frac{x^8)/4+(x^4)/4+C