What is the integral of x^3-3x+\sqrt[4]{x}-5 ?

The integral of x^3-3x+\sqrt[4]{x}-5 is (x^4)/4-(3x^2)/2+4/5 x^{5/4}-5x+C

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integral of x^3-3x+\sqrt[4]{x}-5

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

Solution steps

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 = \frac{x ^{4}}{4}

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

\int 5 d x = 5 x

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

Add a constant to the solution

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

n = 0 \sum \infty 3^{n} 8^{- n + 1}

\frac{d}{d x} (\frac{100 x}{x ^{2} + x + 4})

f (x) = \frac{5}{x}

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

\int_{0}^{π} sin^{2} (3 x) d x

What is the integral of x^3-3x+\sqrt[4]{x}-5 ?
The integral of x^3-3x+\sqrt[4]{x}-5 is (x^4)/4-(3x^2)/2+4/5 x^{5/4}-5x+C