What is the integral of (x+9)/(sqrt(16-(x-4)^2)) ?

The integral of (x+9)/(sqrt(16-(x-4)^2)) is -sqrt(-x^2+8x)+13arcsin(1/4 x-1)+C

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integral of (x+9)/(sqrt(16-(x-4)^2))

\int \frac{x + 9}{16 - ( x - 4 ) ^{2}} d x

- - x^{2} + 8 x + 13 arcsin (\frac{1}{4} x - 1) + C

Solution steps

\int \frac{x + 9}{16 - ( x - 4 ) ^{2}} d x

Apply u-substitution

: \int \frac{u + 13}{16 - u ^{2}} d u

= \int \frac{u + 13}{16 - u ^{2}} d u

Apply Trigonometric Substitution

: \int 4 sin (v) + 13 d v

= \int 4 sin (v) + 13 d v

Apply the Sum Rule:

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

= \int 4 sin (v) d v + \int 13 d v

\int 4 sin (v) d v = - 4 cos (v)

\int 13 d v = 13 v

= - 4 cos (v) + 13 v

Substitute back

= - 4 cos (arcsin (\frac{1}{4} (x - 4))) + 13 arcsin (\frac{1}{4} (x - 4))

Simplify

- 4 cos (arcsin (\frac{1}{4} (x - 4))) + 13 arcsin (\frac{1}{4} (x - 4)) : - - x^{2} + 8 x + 13 arcsin (\frac{1}{4} x - 1)

= - - x^{2} + 8 x + 13 arcsin (\frac{1}{4} x - 1)

Add a constant to the solution

= - - x^{2} + 8 x + 13 arcsin (\frac{1}{4} x - 1) + C

\int (sin (x))^{11} (cos (x))^{5} d x

y^{^{'}} = y^{3} - 6 x^{2} y^{3}, y (1) = - \frac{1}{2}

x \to 1 lim (\frac{x ^{2} + 1}{x + 1})

f (x) = 80640 (2 x + 5)^{2}

f (x) = x^{2} - x

What is the integral of (x+9)/(sqrt(16-(x-4)^2)) ?
The integral of (x+9)/(sqrt(16-(x-4)^2)) is -sqrt(-x^2+8x)+13arcsin(1/4 x-1)+C