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

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

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

\int (2 x - 3 x - x^{4}) d x

\frac{4}{3} x^{\frac{3}{2}} - \frac{3}{4} x^{\frac{4}{3}} - \frac{x ^{5}}{5} + C

Solution steps

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

\int 2 x d x = \frac{4}{3} x^{\frac{3}{2}}

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

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

= \frac{4}{3} x^{\frac{3}{2}} - \frac{3}{4} x^{\frac{4}{3}} - \frac{x ^{5}}{5}

Add a constant to the solution

= \frac{4}{3} x^{\frac{3}{2}} - \frac{3}{4} x^{\frac{4}{3}} - \frac{x ^{5}}{5} + C

y^{^{''}} + y^{^{'}} = sec (x) tan (x)

d er i v a t i v e f (x) = \frac{8 x}{8 + x}

y^{^{'}} = \frac{( 1 + y ^{6} )}{x y ^{5}}

\frac{d}{d x} (x^{3} y^{2} - 3 x sin (x y))

\frac{d}{d x} (\frac{1}{2 x} + 5 cos (x))

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