What is the integral of (5x+3)/(sqrt(5+4x-x^2)) ?

The integral of (5x+3)/(sqrt(5+4x-x^2)) is -5sqrt(-x^2+4x+5)+13arcsin(1/3 (x-2))+C

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integral of (5x+3)/(sqrt(5+4x-x^2))

\int \frac{5 x + 3}{5 + 4 x - x ^{2}} d x

- 5 - x^{2} + 4 x + 5 + 13 arcsin (\frac{1}{3} (x - 2)) + C

Solution steps

\int \frac{5 x + 3}{5 + 4 x - x ^{2}} d x

Complete the square

5 + 4 x - x^{2} : - (x - 2)^{2} + 9

= \int \frac{5 x + 3}{- ( x - 2 ) ^{2} + 9} d x

Apply u-substitution

: \int \frac{5 u + 13}{- u ^{2} + 9} d u

= \int \frac{5 u + 13}{- u ^{2} + 9} d u

Expand

\frac{5 u + 13}{- u ^{2} + 9} : \frac{5 u}{- u ^{2} + 9} + \frac{13}{- u ^{2} + 9}

= \int \frac{5 u}{- u ^{2} + 9} + \frac{13}{- 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{5 u}{- u ^{2} + 9} d u + \int \frac{13}{- u ^{2} + 9} d u

\int \frac{5 u}{- u ^{2} + 9} d u = - 5 - u^{2} + 9

\int \frac{13}{- u ^{2} + 9} d u = 13 arcsin (\frac{1}{3} u)

= - 5 - u^{2} + 9 + 13 arcsin (\frac{1}{3} u)

Substitute back

u = x - 2

= - 5 - (x - 2)^{2} + 9 + 13 arcsin (\frac{1}{3} (x - 2))

Expand

- (x - 2)^{2} + 9 : - x^{2} + 4 x + 5

= - 5 - x^{2} + 4 x + 5 + 13 arcsin (\frac{1}{3} (x - 2))

Add a constant to the solution

= - 5 - x^{2} + 4 x + 5 + 13 arcsin (\frac{1}{3} (x - 2)) + C

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What is the integral of (5x+3)/(sqrt(5+4x-x^2)) ?
The integral of (5x+3)/(sqrt(5+4x-x^2)) is -5sqrt(-x^2+4x+5)+13arcsin(1/3 (x-2))+C