What is the integral of sqrt(y+1) ?

The integral of sqrt(y+1) is 2/3 (y+1)^{3/2}+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of sqrt(y+1)

\int y + 1 d y

\frac{2}{3} (y + 1)^{\frac{3}{2}} + C

Solution steps

\int y + 1 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 + 1

= \frac{2}{3} (y + 1)^{\frac{3}{2}}

Add a constant to the solution

= \frac{2}{3} (y + 1)^{\frac{3}{2}} + C

9 cos (x), 7 sin (x), 0, π

\frac{d}{d x} (x^{2} e^{x} - 2 x e^{x} + 9 e^{x})

\frac{100}{s ( s + 10 ) ^{2}}

y = \frac{4}{x}

x \to 2 π lim (f (x))

What is the integral of sqrt(y+1) ?
The integral of sqrt(y+1) is 2/3 (y+1)^{3/2}+C