What is the integral of ((e^{3x})/3+3x*2^{x^2}-x^4) ?

The integral of ((e^{3x})/3+3x*2^{x^2}-x^4) is 1/9 e^{3x}+(3*2^{x^2-1}}{ln(2)}-\frac{x^5)/5+C

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integral of ((e^{3x})/3+3x*2^{x^2}-x^4)

\int (\frac{e ^{3 x}}{3} + 3 x \cdot 2^{x^{2}} - x^{4}) d x

\frac{1}{9} e^{3 x} + \frac{3 \cdot 2 ^{x^{2} - 1}}{ln ( 2 )} - \frac{x ^{5}}{5} + C

Solution steps

\int \frac{e ^{3 x}}{3} + 3 x \cdot 2^{x^{2}} - x^{4} 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 \frac{e ^{3 x}}{3} d x + \int 3 x \cdot 2^{x^{2}} d x - \int x^{4} d x

\int \frac{e ^{3 x}}{3} d x = \frac{1}{9} e^{3 x}

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

\int x^{4} d x = \frac{x ^{5}}{5}

= \frac{1}{9} e^{3 x} + \frac{3 \cdot 2 ^{x^{2} - 1}}{ln ( 2 )} - \frac{x ^{5}}{5}

Add a constant to the solution

= \frac{1}{9} e^{3 x} + \frac{3 \cdot 2 ^{x^{2} - 1}}{ln ( 2 )} - \frac{x ^{5}}{5} + C

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What is the integral of ((e^{3x})/3+3x*2^{x^2}-x^4) ?
The integral of ((e^{3x})/3+3x*2^{x^2}-x^4) is 1/9 e^{3x}+(3*2^{x^2-1}}{ln(2)}-\frac{x^5)/5+C