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

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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