What is the integral of 1/(2sqrt(x)+3) ?

The integral of 1/(2sqrt(x)+3) is sqrt(x)+3/2-3/2 ln|2sqrt(x)+3|+C

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integral of 1/(2sqrt(x)+3)

\int \frac{1}{2 x + 3} d x

x + \frac{3}{2} - \frac{3}{2} ln 2 x + 3 + C

Solution steps

\int \frac{1}{2 x + 3} d x

Apply u-substitution

: \int \frac{2 u}{2 u + 3} d u

= \int \frac{2 u}{2 u + 3} d u

Take the constant out:

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

= 2 \cdot \int \frac{u}{2 u + 3} d u

Apply u-substitution

: \int \frac{v - 3}{4 v} d v

= 2 \cdot \int \frac{v - 3}{4 v} d v

Take the constant out:

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

= 2 \cdot \frac{1}{4} \cdot \int \frac{v - 3}{v} d v

Expand

\frac{v - 3}{v} : 1 - \frac{3}{v}

= 2 \cdot \frac{1}{4} \cdot \int 1 - \frac{3}{v} d v

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= 2 \cdot \frac{1}{4} (\int 1 d v - \int \frac{3}{v} d v)

\int 1 d v = v

\int \frac{3}{v} d v = 3 ln ∣ v ∣

= 2 \cdot \frac{1}{4} (v - 3 ln ∣ v ∣)

Substitute back

= 2 \cdot \frac{1}{4} (2 x + 3 - 3 ln 2 x + 3)

Simplify

2 \cdot \frac{1}{4} (2 x + 3 - 3 ln 2 x + 3) : x + \frac{3}{2} - \frac{3}{2} ln 2 x + 3

= x + \frac{3}{2} - \frac{3}{2} ln 2 x + 3

Add a constant to the solution

= x + \frac{3}{2} - \frac{3}{2} ln 2 x + 3 + C

\frac{\partial}{\partial y} (2 x (5 + y)^{- 1})

\frac{sec ( 9 x ^{2} )}{tan ( 2 x ) \cdot cos ( x )}

y^{^{'}} = 3

1 + 6 x^{\frac{3}{2}}

\frac{d}{d x} (\frac{6 e ^{x}}{5 e ^{x} + 7})

What is the integral of 1/(2sqrt(x)+3) ?
The integral of 1/(2sqrt(x)+3) is sqrt(x)+3/2-3/2 ln|2sqrt(x)+3|+C