What is the integral of sqrt(7+6x-x^2) ?

The integral of sqrt(7+6x-x^2) is 8(arcsin(1/4 (x-3))+1/2 sin(2arcsin(1/4 (x-3))))+C

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integral of sqrt(7+6x-x^2)

\int 7 + 6 x - x^{2} d x

8 (arcsin (\frac{1}{4} (x - 3)) + \frac{1}{2} sin (2 arcsin (\frac{1}{4} (x - 3)))) + C

Solution steps

\int 7 + 6 x - x^{2} d x

Complete the square

7 + 6 x - x^{2} : - (x - 3)^{2} + 16

= \int - (x - 3)^{2} + 16 d x

Apply u-substitution

: \int - u^{2} + 16 d u

= \int - u^{2} + 16 d u

Apply Trigonometric Substitution

: \int 16 cos^{2} (v) d v

= \int 16 cos^{2} (v) d v

Take the constant out:

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

= 16 \cdot \int cos^{2} (v) d v

Rewrite using trig identities

= 16 \cdot \int \frac{1 + cos ( 2 v )}{2} d v

Take the constant out:

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

= 16 \cdot \frac{1}{2} \cdot \int 1 + cos (2 v) d v

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= 16 \cdot \frac{1}{2} (\int 1 d v + \int cos (2 v) d v)

\int 1 d v = v

\int cos (2 v) d v = \frac{1}{2} sin (2 v)

= 16 \cdot \frac{1}{2} (v + \frac{1}{2} sin (2 v))

Substitute back

= 16 \cdot \frac{1}{2} (arcsin (\frac{1}{4} (x - 3)) + \frac{1}{2} sin (2 arcsin (\frac{1}{4} (x - 3))))

Simplify

16 \cdot \frac{1}{2} (arcsin (\frac{1}{4} (x - 3)) + \frac{1}{2} sin (2 arcsin (\frac{1}{4} (x - 3)))) : 8 (arcsin (\frac{1}{4} (x - 3)) + \frac{1}{2} sin (2 arcsin (\frac{1}{4} (x - 3))))

= 8 (arcsin (\frac{1}{4} (x - 3)) + \frac{1}{2} sin (2 arcsin (\frac{1}{4} (x - 3))))

Add a constant to the solution

= 8 (arcsin (\frac{1}{4} (x - 3)) + \frac{1}{2} sin (2 arcsin (\frac{1}{4} (x - 3)))) + C

\frac{d y}{d x} = 2 e^{x}

x \to - 3 lim (\frac{( x + 2 )}{x + 3})

\frac{\partial}{\partial x} (x^{2} y + 2 x z + 2 y^{2} + 6 z + 9)

\frac{\partial}{\partial t} (x z t^{2} y + 1)

\int e^{x} 1 + e^{x} d x

What is the integral of sqrt(7+6x-x^2) ?
The integral of sqrt(7+6x-x^2) is 8(arcsin(1/4 (x-3))+1/2 sin(2arcsin(1/4 (x-3))))+C