What is the integral of sqrt(16x) ?

The integral of sqrt(16x) is 8/3 x^{3/2}+C

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integral of sqrt(16x)

\int 16 x d x

\frac{8}{3} x^{\frac{3}{2}} + C

Solution steps

\int 16 x d x

Simplify

16 x : 4 x

= \int 4 x d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 4 \cdot \int x d x

Apply the Power Rule

: \frac{2}{3} x^{\frac{3}{2}}

= 4 \cdot \frac{2}{3} x^{\frac{3}{2}}

Simplify

4 \cdot \frac{2}{3} x^{\frac{3}{2}} : \frac{8}{3} x^{\frac{3}{2}}

= \frac{8}{3} x^{\frac{3}{2}}

Add a constant to the solution

= \frac{8}{3} x^{\frac{3}{2}} + C

\int \frac{40 x}{5 x ^{2} + e} d x

x \to 0 + lim (tan (x) ln (x))

\frac{d}{d x} (3 x + x)

y^{^{''}} + 4 y^{^{'}} + 5 y = 5 x + 3 e - x

x \to 1 lim (h (x))