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

The integral of (1/2 x^4+1/4 x^3-x) is (x^5}{10}+(x^4)/(16)-\frac{x^2)/2+C

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

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

\frac{x ^{5}}{10} + \frac{x ^{4}}{16} - \frac{x ^{2}}{2} + C

Solution steps

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

\int \frac{1}{2} x^{4} d x = \frac{x ^{5}}{10}

\int \frac{1}{4} x^{3} d x = \frac{x ^{4}}{16}

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

= \frac{x ^{5}}{10} + \frac{x ^{4}}{16} - \frac{x ^{2}}{2}

Add a constant to the solution

= \frac{x ^{5}}{10} + \frac{x ^{4}}{16} - \frac{x ^{2}}{2} + C

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What is the integral of (1/2 x^4+1/4 x^3-x) ?
The integral of (1/2 x^4+1/4 x^3-x) is (x^5}{10}+(x^4)/(16)-\frac{x^2)/2+C