What is the integral of (6x+12)/(sqrt(x^2+4x-1)) ?

The integral of (6x+12)/(sqrt(x^2+4x-1)) is 6sqrt(x^2+4x-1)+C

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integral of (6x+12)/(sqrt(x^2+4x-1))

\int \frac{6 x + 12}{x ^{2} + 4 x - 1} d x

6 x^{2} + 4 x - 1 + C

Solution steps

\int \frac{6 x + 12}{x ^{2} + 4 x - 1} d x

Apply u-substitution

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

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

Take the constant out:

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

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

Apply the Power Rule

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

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

Substitute back

u = x^{2} + 4 x - 1

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

Simplify

3 \cdot 2 (x^{2} + 4 x - 1)^{\frac{1}{2}} : 6 x^{2} + 4 x - 1

= 6 x^{2} + 4 x - 1

Add a constant to the solution

= 6 x^{2} + 4 x - 1 + C

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What is the integral of (6x+12)/(sqrt(x^2+4x-1)) ?
The integral of (6x+12)/(sqrt(x^2+4x-1)) is 6sqrt(x^2+4x-1)+C