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

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

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

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

2 x + 1 + C

Solution steps

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

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 = x + 1

= 2 x + 1

Add a constant to the solution

= 2 x + 1 + C

f (x) = 8 + 4 x^{2} - 2 x^{3}, at x = 2

x^{3} y^{^{'}} = x^{2} y - y^{3}

\int cos^{3} (\frac{x}{13}) d x

x \to 0 lim (x + 6)

y = (9 x^{4} + 5 x)^{3}

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