What is the integral of (3x+6)/(sqrt(2x^2+8x+3)) ?

The integral of (3x+6)/(sqrt(2x^2+8x+3)) is 3/2 sqrt(2x^2+8x+3)+C

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integral of (3x+6)/(sqrt(2x^2+8x+3))

\int \frac{3 x + 6}{2 x ^{2} + 8 x + 3} d x

\frac{3}{2} 2 x^{2} + 8 x + 3 + C

Solution steps

\int \frac{3 x + 6}{2 x ^{2} + 8 x + 3} d x

Apply u-substitution

: \int \frac{3}{4 u} d u

= \int \frac{3}{4 u} d u

Take the constant out:

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

= \frac{3}{4} \cdot \int \frac{1}{u} d u

Apply the Power Rule

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

= \frac{3}{4} \cdot 2 u^{\frac{1}{2}}

Substitute back

u = 2 x^{2} + 8 x + 3

= \frac{3}{4} \cdot 2 (2 x^{2} + 8 x + 3)^{\frac{1}{2}}

Simplify

\frac{3}{4} \cdot 2 (2 x^{2} + 8 x + 3)^{\frac{1}{2}} : \frac{3}{2} 2 x^{2} + 8 x + 3

= \frac{3}{2} 2 x^{2} + 8 x + 3

Add a constant to the solution

= \frac{3}{2} 2 x^{2} + 8 x + 3 + C

x \to \frac{π}{4} lim (2 cot (2 x))

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

x \to 7 lim (\frac{x - 8}{x - 7})

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

\frac{d}{d x} (\frac{5}{3} x^{3} - x^{2} + 1)

What is the integral of (3x+6)/(sqrt(2x^2+8x+3)) ?
The integral of (3x+6)/(sqrt(2x^2+8x+3)) is 3/2 sqrt(2x^2+8x+3)+C