What is the integral of 3\sqrt[4]{x^3}-8x^5+6e^x-2 ?

The integral of 3\sqrt[4]{x^3}-8x^5+6e^x-2 is 12/7 x^{7/4}-(4x^6)/3+6e^x-2x+C

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integral of 3\sqrt[4]{x^3}-8x^5+6e^x-2

\frac{12}{7} x^{\frac{7}{4}} - \frac{4 x ^{6}}{3} + 6 e^{x} - 2 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 8 x^{5} d x = \frac{4 x ^{6}}{3}

\int 6 e^{x} d x = 6 e^{x}

\int 2 d x = 2 x

= \frac{12}{7} x^{\frac{7}{4}} - \frac{4 x ^{6}}{3} + 6 e^{x} - 2 x

Add a constant to the solution

= \frac{12}{7} x^{\frac{7}{4}} - \frac{4 x ^{6}}{3} + 6 e^{x} - 2 x + C

What is the integral of 3\sqrt[4]{x^3}-8x^5+6e^x-2 ?
The integral of 3\sqrt[4]{x^3}-8x^5+6e^x-2 is 12/7 x^{7/4}-(4x^6)/3+6e^x-2x+C