What is the integral of (2x+5)/(sqrt(-x^2-6x+16)) ?

The integral of (2x+5)/(sqrt(-x^2-6x+16)) is -2sqrt(-x^2-6x+16)-arcsin(1/5 (x+3))+C

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integral of (2x+5)/(sqrt(-x^2-6x+16))

\int \frac{2 x + 5}{- x ^{2} - 6 x + 16} d x

- 2 - x^{2} - 6 x + 16 - arcsin (\frac{1}{5} (x + 3)) + C

Solution steps

\int \frac{2 x + 5}{- x ^{2} - 6 x + 16} d x

Complete the square

- x^{2} - 6 x + 16 : - (x + 3)^{2} + 25

= \int \frac{2 x + 5}{- ( x + 3 ) ^{2} + 25} d x

Apply u-substitution

: \int \frac{2 u - 1}{- u ^{2} + 25} d u

= \int \frac{2 u - 1}{- u ^{2} + 25} d u

Expand

\frac{2 u - 1}{- u ^{2} + 25} : \frac{2 u}{- u ^{2} + 25} - \frac{1}{- u ^{2} + 25}

= \int \frac{2 u}{- u ^{2} + 25} - \frac{1}{- u ^{2} + 25} 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{2 u}{- u ^{2} + 25} d u - \int \frac{1}{- u ^{2} + 25} d u

\int \frac{2 u}{- u ^{2} + 25} d u = - 2 - u^{2} + 25

\int \frac{1}{- u ^{2} + 25} d u = arcsin (\frac{1}{5} u)

= - 2 - u^{2} + 25 - arcsin (\frac{1}{5} u)

Substitute back

u = x + 3

= - 2 - (x + 3)^{2} + 25 - arcsin (\frac{1}{5} (x + 3))

Expand

- (x + 3)^{2} + 25 : - x^{2} - 6 x + 16

= - 2 - x^{2} - 6 x + 16 - arcsin (\frac{1}{5} (x + 3))

Add a constant to the solution

= - 2 - x^{2} - 6 x + 16 - arcsin (\frac{1}{5} (x + 3)) + C

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What is the integral of (2x+5)/(sqrt(-x^2-6x+16)) ?
The integral of (2x+5)/(sqrt(-x^2-6x+16)) is -2sqrt(-x^2-6x+16)-arcsin(1/5 (x+3))+C