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

y = \frac{1}{5 + 2 x}, (- 1, \frac{1}{3})

x \frac{d y}{d x} = x^{2} - y^{2} + y

\int - t d t

x \to 3 lim (x (\frac{2}{x} + 5))

\int_{\frac{1}{7}}^{5} 12 x ln (7 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