What is the integral of 2\sqrt[5]{x^4}-7x^3+10e^x-1 ?

The integral of 2\sqrt[5]{x^4}-7x^3+10e^x-1 is 10/9 x^{9/5}-(7x^4)/4+10e^x-x+C

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integral of 2\sqrt[5]{x^4}-7x^3+10e^x-1

\frac{10}{9} x^{\frac{9}{5}} - \frac{7 x ^{4}}{4} + 10 e^{x} - x + C

Solution steps

assuming

x \geq 0

= \int 2 x^{\frac{4}{5}} - 7 x^{3} + 10 e^{x} - 1 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 2 x^{\frac{4}{5}} d x - \int 7 x^{3} d x + \int 10 e^{x} d x - \int 1 d x

\int 2 x^{\frac{4}{5}} d x = \frac{10}{9} x^{\frac{9}{5}}

\int 7 x^{3} d x = \frac{7 x ^{4}}{4}

\int 10 e^{x} d x = 10 e^{x}

\int 1 d x = x

= \frac{10}{9} x^{\frac{9}{5}} - \frac{7 x ^{4}}{4} + 10 e^{x} - x

Add a constant to the solution

= \frac{10}{9} x^{\frac{9}{5}} - \frac{7 x ^{4}}{4} + 10 e^{x} - x + C

x^{x^{3} - 2 x}

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

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

\int_{- 6}^{1} - 2 x^{2} - 10 x + 12 d x

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

What is the integral of 2\sqrt[5]{x^4}-7x^3+10e^x-1 ?
The integral of 2\sqrt[5]{x^4}-7x^3+10e^x-1 is 10/9 x^{9/5}-(7x^4)/4+10e^x-x+C