What is the integral of (x-1)^2*(x+1)^2 ?

The integral of (x-1)^2*(x+1)^2 is -x^5+(6x^5)/5-(2x^3)/3+x+C

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integral of (x-1)^2*(x+1)^2

\int (x - 1)^{2} \cdot (x + 1)^{2} d x

- x^{5} + \frac{6 x ^{5}}{5} - \frac{2 x ^{3}}{3} + x + C

Solution steps

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

Apply Integration By Parts

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

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

\int (4 x^{3} - 4 x) x d x = 2 x^{3} (x^{2} - 1) - 2 (\frac{3 x ^{5}}{5} - \frac{x ^{3}}{3})

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

Simplify

(x - 1)^{2} (x + 1)^{2} x - (2 x^{3} (x^{2} - 1) - 2 (\frac{3 x ^{5}}{5} - \frac{x ^{3}}{3})) : - x^{5} + \frac{6 x ^{5}}{5} - \frac{2 x ^{3}}{3} + x

= - x^{5} + \frac{6 x ^{5}}{5} - \frac{2 x ^{3}}{3} + x

Add a constant to the solution

= - x^{5} + \frac{6 x ^{5}}{5} - \frac{2 x ^{3}}{3} + x + C

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What is the integral of (x-1)^2*(x+1)^2 ?
The integral of (x-1)^2*(x+1)^2 is -x^5+(6x^5)/5-(2x^3)/3+x+C