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

The integral of (-12x-1)sqrt(-6x^2-x+1) is 1/(72sqrt(6))(-144x^2-24x+24)^{3/2}+C

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

\int (- 12 x - 1) - 6 x^{2} - x + 1 d x

\frac{1}{72 6} (- 144 x^{2} - 24 x + 24)^{\frac{3}{2}} + C

Solution steps

\int (- 12 x - 1) - 6 x^{2} - x + 1 d x

Apply u-substitution

: \int - \frac{u - u ^{2} + 25}{24 6} d u

= \int - \frac{u - u ^{2} + 25}{24 6} d u

Take the constant out:

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

= - \frac{1}{24 6} \cdot \int u - u^{2} + 25 d u

Apply u-substitution

: \int - \frac{v}{2} d v

= - \frac{1}{24 6} \cdot \int - \frac{v}{2} d v

Take the constant out:

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

= - \frac{1}{24 6} (- \frac{1}{2} \cdot \int v d v)

Apply the Power Rule

: \frac{2}{3} v^{\frac{3}{2}}

= - \frac{1}{24 6} (- \frac{1}{2} \cdot \frac{2}{3} v^{\frac{3}{2}})

Substitute back

= - \frac{1}{24 6} (- \frac{1}{2} \cdot \frac{2}{3} (- (- 12 x - 1)^{2} + 25)^{\frac{3}{2}})

Simplify

- \frac{1}{24 6} (- \frac{1}{2} \cdot \frac{2}{3} (- (- 12 x - 1)^{2} + 25)^{\frac{3}{2}}) : \frac{1}{72 6} (- 144 x^{2} - 24 x + 24)^{\frac{3}{2}}

= \frac{1}{72 6} (- 144 x^{2} - 24 x + 24)^{\frac{3}{2}}

Add a constant to the solution

= \frac{1}{72 6} (- 144 x^{2} - 24 x + 24)^{\frac{3}{2}} + C

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What is the integral of (-12x-1)sqrt(-6x^2-x+1) ?
The integral of (-12x-1)sqrt(-6x^2-x+1) is 1/(72sqrt(6))(-144x^2-24x+24)^{3/2}+C