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

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

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

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

2 3 + 2 x + C

Solution steps

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

Take the constant out:

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

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

Apply u-substitution

: \int \frac{1}{2 u} d u

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

Take the constant out:

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

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

Apply the Power Rule

: 2 u^{\frac{1}{2}}

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

Substitute back

u = 3 + 2 x

= 2 \cdot \frac{1}{2} \cdot 2 (3 + 2 x)^{\frac{1}{2}}

Simplify

2 \cdot \frac{1}{2} \cdot 2 (3 + 2 x)^{\frac{1}{2}} : 2 3 + 2 x

= 2 3 + 2 x

Add a constant to the solution

= 2 3 + 2 x + C

x \to 3 lim (\frac{x}{x - 3})

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

\int_{1}^{2} 5 x^{2} - 4 x + 3 d x

\int \frac{1}{x ^{2} + 10} d x

x^{2} y^{^{''}} + x y^{^{'}} + 9 y = 0, y (1) = 0, y^{^{'}} (1) = 2.5

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