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

The integral of (1/5 x^5+1/3 x^4-x) is (x^6}{30}+(x^5)/(15)-\frac{x^2)/2+C

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integral of (1/5 x^5+1/3 x^4-x)

\int (\frac{1}{5} x^{5} + \frac{1}{3} x^{4} - x) d x

\frac{x ^{6}}{30} + \frac{x ^{5}}{15} - \frac{x ^{2}}{2} + C

Solution steps

\int \frac{1}{5} x^{5} + \frac{1}{3} x^{4} - 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}{5} x^{5} d x + \int \frac{1}{3} x^{4} d x - \int x d x

\int \frac{1}{5} x^{5} d x = \frac{x ^{6}}{30}

\int \frac{1}{3} x^{4} d x = \frac{x ^{5}}{15}

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

= \frac{x ^{6}}{30} + \frac{x ^{5}}{15} - \frac{x ^{2}}{2}

Add a constant to the solution

= \frac{x ^{6}}{30} + \frac{x ^{5}}{15} - \frac{x ^{2}}{2} + C

x \to 0 lim ((1 - 2 x)^{(\frac{1}{x})})

\frac{d}{d x} (ln (10 x^{2} - 8))

y^{^{'}} = e^{2 t + y - 1} - 2

\frac{\partial}{\partial x} (e^{x y} + ln (\frac{x}{y}))

\frac{\partial}{\partial x} (2 x y^{2} + 4)

What is the integral of (1/5 x^5+1/3 x^4-x) ?
The integral of (1/5 x^5+1/3 x^4-x) is (x^6}{30}+(x^5)/(15)-\frac{x^2)/2+C