What is the integral of 1/(sqrt(5-8x-x^2)) ?

The integral of 1/(sqrt(5-8x-x^2)) is arcsin(1/(sqrt(21))(x+4))+C

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integral of 1/(sqrt(5-8x-x^2))

\int \frac{1}{5 - 8 x - x ^{2}} d x

arcsin (\frac{1}{21} (x + 4)) + C

Solution steps

\int \frac{1}{5 - 8 x - x ^{2}} d x

Complete the square

5 - 8 x - x^{2} : - (x + 4)^{2} + 21

= \int \frac{1}{- ( x + 4 ) ^{2} + 21} d x

Apply u-substitution

: \int \frac{1}{- u ^{2} + 21} d u

= \int \frac{1}{- u ^{2} + 21} 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}{21} (x + 4))

Simplify

= arcsin (\frac{1}{21} (x + 4))

Add a constant to the solution

= arcsin (\frac{1}{21} (x + 4)) + C

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

\frac{d}{d x} (ln (4 - 3 x^{3}))

\frac{- 2 x}{( x ^{2} + 1 ) ^{2}}

t \cdot y^{^{'}} + 2 \cdot y = t^{2} - t + 1

f (x) = e^{- 3 x} sin (4 x^{2})

What is the integral of 1/(sqrt(5-8x-x^2)) ?
The integral of 1/(sqrt(5-8x-x^2)) is arcsin(1/(sqrt(21))(x+4))+C