What is the integral of (x^4+2sqrt(x)-16) ?

The integral of (x^4+2sqrt(x)-16) is (x^5)/5+4/3 x^{3/2}-16x+C

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integral of (x^4+2sqrt(x)-16)

\int (x^{4} + 2 x - 16) d x

\frac{x ^{5}}{5} + \frac{4}{3} x^{\frac{3}{2}} - 16 x + C

Solution steps

\int x^{4} + 2 x - 16 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^{4} d x + \int 2 x d x - \int 16 d x

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

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

\int 16 d x = 16 x

= \frac{x ^{5}}{5} + \frac{4}{3} x^{\frac{3}{2}} - 16 x

Add a constant to the solution

= \frac{x ^{5}}{5} + \frac{4}{3} x^{\frac{3}{2}} - 16 x + C

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\frac{d}{d x} (a (1 - cos (x)))

\frac{\partial}{\partial x} (e^{2 y^{2}} + 4 x^{2})

\int_{1}^{5} \frac{1}{( x - 4 ) ^{2}} d x

\int - 2 x cos (3 x) d x

What is the integral of (x^4+2sqrt(x)-16) ?
The integral of (x^4+2sqrt(x)-16) is (x^5)/5+4/3 x^{3/2}-16x+C