What is the integral from-x to x^2+1 of (x^3-x) ?

The integral from-x to x^2+1 of (x^3-x) is (x^8+4x^6+5x^4+4x^2+1)/4-(x^4+x^2+1)/2

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integral from-x to x^2+1 of (x^3-x)

\int_{- x}^{x^{2} + 1} (x^{3} - x) d x

\frac{x ^{8} + 4 x ^{6} + 5 x ^{4} + 4 x ^{2} + 1}{4} - \frac{x ^{4} + x ^{2} + 1}{2}

Solution steps

\int_{- x}^{x^{2} + 1} x^{3} - x 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}^{x^{2} + 1} x^{3} d x - \int_{- x}^{x^{2} + 1} x d x

\int_{- x}^{x^{2} + 1} x^{3} d x = \frac{x ^{8} + 4 x ^{6} + 5 x ^{4} + 4 x ^{2} + 1}{4}

\int_{- x}^{x^{2} + 1} x d x = \frac{x ^{4} + x ^{2} + 1}{2}

= \frac{x ^{8} + 4 x ^{6} + 5 x ^{4} + 4 x ^{2} + 1}{4} - \frac{x ^{4} + x ^{2} + 1}{2}

\int_{- 1}^{2} (x^{3} - 2 x^{2} - 5 x + 6) d x

\int_{0}^{3} x 1 + x^{2} d x

\int_{2}^{3} sin (\frac{π}{3} x) d x

\int_{0}^{1} y sin (x y) d x

\int_{0}^{l n (20)} tanh (x) d x

What is the integral from-x to x^2+1 of (x^3-x) ?
The integral from-x to x^2+1 of (x^3-x) is (x^8+4x^6+5x^4+4x^2+1)/4-(x^4+x^2+1)/2