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

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

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integral of 1/(2sqrt(y+1))

\int \frac{1}{2 y + 1} d y

y + 1 + C

Solution steps

\int \frac{1}{2 y + 1} d y

Take the constant out:

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

= \frac{1}{2} \cdot \int \frac{1}{y + 1} d y

Apply u-substitution

: \int 2 d u

= \frac{1}{2} \cdot \int 2 d u

Integral of a constant:

\int a d x = a x

= \frac{1}{2} \cdot 2 u

Substitute back

u = y + 1

= \frac{1}{2} \cdot 2 y + 1

Simplify

\frac{1}{2} \cdot 2 y + 1 : y + 1

= y + 1

Add a constant to the solution

= y + 1 + C

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