What is the integral of 1/(sqrt(3-8x-x^2)) ?

The integral of 1/(sqrt(3-8x-x^2)) is arcsin(1/(sqrt(19))(x+4))+C

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integral of 1/(sqrt(3-8x-x^2))

\int \frac{1}{3 - 8 x - x ^{2}} d x

arcsin (\frac{1}{19} (x + 4)) + C

Solution steps

\int \frac{1}{3 - 8 x - x ^{2}} d x

Complete the square

3 - 8 x - x^{2} : - (x + 4)^{2} + 19

= \int \frac{1}{- ( x + 4 ) ^{2} + 19} d x

Apply u-substitution

: \int \frac{1}{- u ^{2} + 19} d u

= \int \frac{1}{- u ^{2} + 19} 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}{19} (x + 4))

Simplify

= arcsin (\frac{1}{19} (x + 4))

Add a constant to the solution

= arcsin (\frac{1}{19} (x + 4)) + C

x \to 1 lim (\frac{x - 11}{x + 7})

\int \frac{1}{u 6 - u ^{2}} d u

\frac{d}{d x} (\frac{x ^{4}}{4} + \frac{1}{8 x ^{2}})

\int_{0}^{2} x^{2} ln (x) d x

f (x) = x^{3} + 1, (1, 2)

What is the integral of 1/(sqrt(3-8x-x^2)) ?
The integral of 1/(sqrt(3-8x-x^2)) is arcsin(1/(sqrt(19))(x+4))+C