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

The integral of 1/(sqrt(y+1)) is 2sqrt(y+1)+C

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integral of 1/(sqrt(y+1))

\int \frac{1}{y + 1} d y

2 y + 1 + C

Solution steps

\int \frac{1}{y + 1} d y

Apply u-substitution

: \int 2 d u

= \int 2 d u

Integral of a constant:

\int a d x = a x

= 2 u

Substitute back

u = y + 1

= 2 y + 1

Add a constant to the solution

= 2 y + 1 + C

\frac{sin ( t )}{4 + 4 cos ( t )}

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

\int_{0}^{\frac{π}{2}} (sin (x))^{2} d x

e^{- 2 t} \cdot sin (t)

\frac{\partial}{\partial y} (y sin (z))

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