What is the integral of 1/(sqrt(-7-8x-x^2)) ?

The integral of 1/(sqrt(-7-8x-x^2)) is arcsin(1/3 (x+4))+C

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integral of 1/(sqrt(-7-8x-x^2))

\int \frac{1}{- 7 - 8 x - x ^{2}} d x

arcsin (\frac{1}{3} (x + 4)) + C

Solution steps

\int \frac{1}{- 7 - 8 x - x ^{2}} d x

Complete the square

- 7 - 8 x - x^{2} : - (x + 4)^{2} + 9

= \int \frac{1}{- ( x + 4 ) ^{2} + 9} d x

Apply u-substitution

: \int \frac{1}{- u ^{2} + 9} d u

= \int \frac{1}{- u ^{2} + 9} 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}{3} (x + 4))

Simplify

= arcsin (\frac{1}{3} (x + 4))

Add a constant to the solution

= arcsin (\frac{1}{3} (x + 4)) + C

\frac{d}{d x} (ln (x) + ln (y))

\frac{\partial}{\partial x} (y^{2} z ln (1 + x^{2}))

y = cos^{l n (7 x)} (x)

\int \frac{1}{sin ( x ) - cos ( x ) - 1} d x

\int_{0}^{\frac{7}{3}} 9 + 3 x d x

What is the integral of 1/(sqrt(-7-8x-x^2)) ?
The integral of 1/(sqrt(-7-8x-x^2)) is arcsin(1/3 (x+4))+C