What is the integral of (4y)/(sqrt(-y^2-8y-7)) ?

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

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integral of (4y)/(sqrt(-y^2-8y-7))

\int \frac{4 y}{- y ^{2} - 8 y - 7} d y

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

Solution steps

\int \frac{4 y}{- y ^{2} - 8 y - 7} d y

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 4 \cdot \int \frac{y}{- y ^{2} - 8 y - 7} d y

Complete the square

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

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

Apply u-substitution

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

= 4 \cdot \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}

= 4 \cdot \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

= 4 (\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)

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

Substitute back

u = y + 4

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

Expand

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

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

Add a constant to the solution

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

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