What is the integral of (3-x)/(sqrt(16+6x-x^2)) ?

The integral of (3-x)/(sqrt(16+6x-x^2)) is sqrt(16+6x-x^2)+C

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integral of (3-x)/(sqrt(16+6x-x^2))

\int \frac{3 - x}{16 + 6 x - x ^{2}} d x

16 + 6 x - x^{2} + C

Solution steps

\int \frac{3 - x}{16 + 6 x - x ^{2}} d x

Apply u-substitution

: \int \frac{1}{2 u} d u

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

Take the constant out:

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

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

Apply the Power Rule

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

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

Substitute back

u = 16 + 6 x - x^{2}

= \frac{1}{2} \cdot 2 (16 + 6 x - x^{2})^{\frac{1}{2}}

Simplify

\frac{1}{2} \cdot 2 (16 + 6 x - x^{2})^{\frac{1}{2}} : 16 + 6 x - x^{2}

= 16 + 6 x - x^{2}

Add a constant to the solution

= 16 + 6 x - x^{2} + C

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What is the integral of (3-x)/(sqrt(16+6x-x^2)) ?
The integral of (3-x)/(sqrt(16+6x-x^2)) is sqrt(16+6x-x^2)+C