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

The integral of (x^4-2x^3+3x^2-1) is (x^5)/5-(x^4)/2+x^3-x+C

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

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

\frac{x ^{5}}{5} - \frac{x ^{4}}{2} + x^{3} - x + C

Solution steps

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

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

\int 2 x^{3} d x = \frac{x ^{4}}{2}

\int 3 x^{2} d x = x^{3}

\int 1 d x = x

= \frac{x ^{5}}{5} - \frac{x ^{4}}{2} + x^{3} - x

Add a constant to the solution

= \frac{x ^{5}}{5} - \frac{x ^{4}}{2} + x^{3} - x + C

What is the integral of (x^4-2x^3+3x^2-1) ?
The integral of (x^4-2x^3+3x^2-1) is (x^5)/5-(x^4)/2+x^3-x+C