What is the integral of sqrt(4+x) ?

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

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integral of sqrt(4+x)

\int 4 + x d x

\frac{2}{3} (4 + x)^{\frac{3}{2}} + C

Solution steps

\int 4 + x d x

Apply u-substitution

: \int u d u

= \int u d u

Apply the Power Rule

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

= \frac{2}{3} u^{\frac{3}{2}}

Substitute back

u = 4 + x

= \frac{2}{3} (4 + x)^{\frac{3}{2}}

Add a constant to the solution

= \frac{2}{3} (4 + x)^{\frac{3}{2}} + C

\frac{d}{d x} (x^{2} + 2 x)

(y - x y^{2}) d x + x d y = 0

n = 1 \sum \infty \frac{1}{n ^{2} + n}

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

f (x) = \frac{10}{x ^{9}}

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