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

The integral of 1/(sqrt(n+1)) is x/(sqrt(n+1))+C

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

\int \frac{1}{n + 1}

\frac{x}{n + 1} + C

Solution steps

\int \frac{1}{n + 1} d x

Integral of a constant:

\int a d x = a x

= \frac{1}{n + 1} x

Simplify

\frac{1}{n + 1} x : \frac{x}{n + 1}

= \frac{x}{n + 1}

Add a constant to the solution

= \frac{x}{n + 1} + C

(6 x^{3} + 2 x + 9) (x^{2} + 6)

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

\frac{\partial}{\partial x} (2 e^{x})

x y^{^{'}} + y = 2 x y^{2}

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

What is the integral of 1/(sqrt(n+1)) ?
The integral of 1/(sqrt(n+1)) is x/(sqrt(n+1))+C