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

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

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integral of sqrt(y+4)

\int y + 4 d y

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

Solution steps

\int y + 4 d y

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 = y + 4

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

Add a constant to the solution

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

\frac{d}{d x} (\frac{2 x}{x ^{4} + 1})

\int_{1}^{4} - 10 t ln (t) d t

\int 10 x^{5} e^{- x^{3}} d x

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

A x + B

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