What is the integral of 1/(sqrt(16+6x-x^2)) ?

The integral of 1/(sqrt(16+6x-x^2)) is arcsin(1/5 (x-3))+C

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integral of 1/(sqrt(16+6x-x^2))

\int \frac{1}{16 + 6 x - x ^{2}} d x

arcsin (\frac{1}{5} (x - 3)) + C

Solution steps

\int \frac{1}{16 + 6 x - x ^{2}} d x

Complete the square

16 + 6 x - x^{2} : - (x - 3)^{2} + 25

= \int \frac{1}{- ( x - 3 ) ^{2} + 25} d x

Apply u-substitution

: \int \frac{1}{- u ^{2} + 25} d u

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

Apply Trigonometric Substitution

: \int 1 d v

= \int 1 d v

Integral of a constant:

\int a d x = a x

= 1 \cdot v

Substitute back

= 1 \cdot arcsin (\frac{1}{5} (x - 3))

Simplify

= arcsin (\frac{1}{5} (x - 3))

Add a constant to the solution

= arcsin (\frac{1}{5} (x - 3)) + C

\frac{\partial}{\partial x} (\frac{y ^{4}}{x})

(2 t - 1) (3 t - 3)^{- 1}

\frac{d}{d x} (arcsin (x - y))

\int_{- 1}^{3} 3 x + 1 d x

y = x^{2}, at x = \frac{1}{3}

What is the integral of 1/(sqrt(16+6x-x^2)) ?
The integral of 1/(sqrt(16+6x-x^2)) is arcsin(1/5 (x-3))+C