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

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

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

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

- - x^{2} - 8 x - 7 - 4 arcsin (\frac{1}{3} (x + 4)) + C

Solution steps

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

Complete the square

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

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

Apply u-substitution

: \int \frac{u - 4}{- u ^{2} + 9} d u

= \int \frac{u - 4}{- u ^{2} + 9} d u

Expand

\frac{u - 4}{- u ^{2} + 9} : \frac{u}{- u ^{2} + 9} - \frac{4}{- u ^{2} + 9}

= \int \frac{u}{- u ^{2} + 9} - \frac{4}{- u ^{2} + 9} d u

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int \frac{u}{- u ^{2} + 9} d u - \int \frac{4}{- u ^{2} + 9} d u

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

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

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

Substitute back

u = x + 4

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

Expand

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

= - - x^{2} - 8 x - 7 - 4 arcsin (\frac{1}{3} (x + 4))

Add a constant to the solution

= - - x^{2} - 8 x - 7 - 4 arcsin (\frac{1}{3} (x + 4)) + C

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