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

The integral of (x^4)/4-(x^3)/3+5x is (x^5}{20}-(x^4)/(12)+\frac{5x^2)/2+C

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integral of (x^4)/4-(x^3)/3+5x

\int \frac{x ^{4}}{4} - \frac{x ^{3}}{3} + 5 x d x

\frac{x ^{5}}{20} - \frac{x ^{4}}{12} + \frac{5 x ^{2}}{2} + C

Solution steps

\int \frac{x ^{4}}{4} - \frac{x ^{3}}{3} + 5 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{x ^{4}}{4} d x - \int \frac{x ^{3}}{3} d x + \int 5 x d x

\int \frac{x ^{4}}{4} d x = \frac{x ^{5}}{20}

\int \frac{x ^{3}}{3} d x = \frac{x ^{4}}{12}

\int 5 x d x = \frac{5 x ^{2}}{2}

= \frac{x ^{5}}{20} - \frac{x ^{4}}{12} + \frac{5 x ^{2}}{2}

Add a constant to the solution

= \frac{x ^{5}}{20} - \frac{x ^{4}}{12} + \frac{5 x ^{2}}{2} + C

\frac{d}{d x} (x^{-})

\frac{4 s}{4 s ^{2} + 1}

\frac{d}{d x} (sin (\frac{π x}{6}))

y = x^{2} e^{x}

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

What is the integral of (x^4)/4-(x^3)/3+5x ?
The integral of (x^4)/4-(x^3)/3+5x is (x^5}{20}-(x^4)/(12)+\frac{5x^2)/2+C