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

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

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

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

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

Solution steps

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

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

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

\int 1 d x = x

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

Add a constant to the solution

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

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