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

The integral of pi(x^3-1)^2-(x-1)^2 is (pix^7)/7-(pix^4)/2+pix-(x^3)/3+x^2-x+C

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

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

\frac{π x ^{7}}{7} - \frac{π x ^{4}}{2} + π x - \frac{x ^{3}}{3} + x^{2} - x + C

Solution steps

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

\int π (x^{3} - 1)^{2} d x = π (\frac{x ^{7}}{7} - \frac{x ^{4}}{2} + x)

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

= π (\frac{x ^{7}}{7} - \frac{x ^{4}}{2} + x) - (\frac{x ^{3}}{3} - x^{2} + x)

Simplify

π (\frac{x ^{7}}{7} - \frac{x ^{4}}{2} + x) - (\frac{x ^{3}}{3} - x^{2} + x) : \frac{π x ^{7}}{7} - \frac{π x ^{4}}{2} + π x - \frac{x ^{3}}{3} + x^{2} - x

= \frac{π x ^{7}}{7} - \frac{π x ^{4}}{2} + π x - \frac{x ^{3}}{3} + x^{2} - x

Add a constant to the solution

= \frac{π x ^{7}}{7} - \frac{π x ^{4}}{2} + π x - \frac{x ^{3}}{3} + x^{2} - x + C

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What is the integral of pi(x^3-1)^2-(x-1)^2 ?
The integral of pi(x^3-1)^2-(x-1)^2 is (pix^7)/7-(pix^4)/2+pix-(x^3)/3+x^2-x+C