What is the integral of 1/(sqrt(u+3)) ?

The integral of 1/(sqrt(u+3)) is 2sqrt(u+3)+C

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integral of 1/(sqrt(u+3))

\int \frac{1}{u + 3} d u

2 u + 3 + C

Solution steps

\int \frac{1}{u + 3} d u

Apply u-substitution

: \int 2 d v

= \int 2 d v

Integral of a constant:

\int a d x = a x

= 2 v

Substitute back

v = u + 3

= 2 u + 3

Add a constant to the solution

= 2 u + 3 + C

\frac{d}{d x} ((x^{4} - 3 x)^{2})

x^{2} y^{^{''}} + 2 x y^{^{'}} - 2 y = x^{2} ln (x)

\frac{1}{4} \cdot sin (2 t)

\frac{d}{d x} (\frac{e ^{x} x - e ^{x}}{7 x ^{2}})

\frac{\partial}{\partial y} (sin (x - y) + (x - y))

What is the integral of 1/(sqrt(u+3)) ?
The integral of 1/(sqrt(u+3)) is 2sqrt(u+3)+C