What is the integral of (\sqrt[5]{x^3}+4^x+14x^7-2) ?

The integral of (\sqrt[5]{x^3}+4^x+14x^7-2) is 5/8 x^{8/5}+(2^{2x-1}}{ln(2)}+\frac{7x^8)/4-2x+C

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integral of (\sqrt[5]{x^3}+4^x+14x^7-2)

\int (5 x^{3} + 4^{x} + 14 x^{7} - 2) d x

\frac{5}{8} x^{\frac{8}{5}} + \frac{2 ^{2 x - 1}}{ln ( 2 )} + \frac{7 x ^{8}}{4} - 2 x + C

Solution steps

\int 5 x^{3} + 4^{x} + 14 x^{7} - 2 d x

5 x^{3} = x^{\frac{3}{5}},

assuming

x \geq 0

= \int x^{\frac{3}{5}} + 4^{x} + 14 x^{7} - 2 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 x^{\frac{3}{5}} d x + \int 4^{x} d x + \int 14 x^{7} d x - \int 2 d x

\int x^{\frac{3}{5}} d x = \frac{5}{8} x^{\frac{8}{5}}

\int 4^{x} d x = \frac{2 ^{2 x - 1}}{ln ( 2 )}

\int 14 x^{7} d x = \frac{7 x ^{8}}{4}

\int 2 d x = 2 x

= \frac{5}{8} x^{\frac{8}{5}} + \frac{2 ^{2 x - 1}}{ln ( 2 )} + \frac{7 x ^{8}}{4} - 2 x

Add a constant to the solution

= \frac{5}{8} x^{\frac{8}{5}} + \frac{2 ^{2 x - 1}}{ln ( 2 )} + \frac{7 x ^{8}}{4} - 2 x + C

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What is the integral of (\sqrt[5]{x^3}+4^x+14x^7-2) ?
The integral of (\sqrt[5]{x^3}+4^x+14x^7-2) is 5/8 x^{8/5}+(2^{2x-1}}{ln(2)}+\frac{7x^8)/4-2x+C