What is the integral of (4x+6)/(sqrt(x^2+3x+1)) ?

The integral of (4x+6)/(sqrt(x^2+3x+1)) is 4sqrt(x^2+3x+1)+C

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integral of (4x+6)/(sqrt(x^2+3x+1))

\int \frac{4 x + 6}{x ^{2} + 3 x + 1} d x

4 x^{2} + 3 x + 1 + C

Solution steps

\int \frac{4 x + 6}{x ^{2} + 3 x + 1} d x

Apply u-substitution

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

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

Take the constant out:

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

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

Apply the Power Rule

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

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

Substitute back

u = x^{2} + 3 x + 1

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

Simplify

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

= 4 x^{2} + 3 x + 1

Add a constant to the solution

= 4 x^{2} + 3 x + 1 + C

y = \frac{( 2 x ^{2} + 7 )}{x ^{2} + 3}

x \to 3 π + lim (csc (x))

f (x) = - \frac{5}{x ^{2}}

4 x y (\frac{d y}{d x}) = 1

\int \frac{5}{4 x ^{2} - 49} d x

What is the integral of (4x+6)/(sqrt(x^2+3x+1)) ?
The integral of (4x+6)/(sqrt(x^2+3x+1)) is 4sqrt(x^2+3x+1)+C