What is the integral of 2x^{1/4}-3x^3 ?

The integral of 2x^{1/4}-3x^3 is 8/5 x^{5/4}-(3x^4)/4+C

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integral of 2x^{1/4}-3x^3

\int 2 x^{\frac{1}{4}} - 3 x^{3} d x

\frac{8}{5} x^{\frac{5}{4}} - \frac{3 x ^{4}}{4} + C

Solution steps

\int 2 x^{\frac{1}{4}} - 3 x^{3} 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{1}{4}} d x - \int 3 x^{3} d x

\int 2 x^{\frac{1}{4}} d x = \frac{8}{5} x^{\frac{5}{4}}

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

= \frac{8}{5} x^{\frac{5}{4}} - \frac{3 x ^{4}}{4}

Add a constant to the solution

= \frac{8}{5} x^{\frac{5}{4}} - \frac{3 x ^{4}}{4} + C

\frac{d}{d θ} (14 sin (θ))

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

\int \frac{1}{a - b y} d y

y^{^{'}} = y (7 - y) - \frac{49}{4}, y (0) = a

y^{^{'}} = \frac{( 3 x ^{4} + y ^{4} )}{x y ^{3}}

What is the integral of 2x^{1/4}-3x^3 ?
The integral of 2x^{1/4}-3x^3 is 8/5 x^{5/4}-(3x^4)/4+C