What is the integral of (x+9)/(sqrt(64-(x-8)^2)) ?

The integral of (x+9)/(sqrt(64-(x-8)^2)) is -sqrt(-x^2+16x)+17arcsin(1/8 x-1)+C

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integral of (x+9)/(sqrt(64-(x-8)^2))

\int \frac{x + 9}{64 - ( x - 8 ) ^{2}} d x

- - x^{2} + 16 x + 17 arcsin (\frac{1}{8} x - 1) + C

Solution steps

\int \frac{x + 9}{64 - ( x - 8 ) ^{2}} d x

Apply u-substitution

: \int \frac{u + 17}{64 - u ^{2}} d u

= \int \frac{u + 17}{64 - u ^{2}} d u

Apply Trigonometric Substitution

: \int 8 sin (v) + 17 d v

= \int 8 sin (v) + 17 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 8 sin (v) d v + \int 17 d v

\int 8 sin (v) d v = - 8 cos (v)

\int 17 d v = 17 v

= - 8 cos (v) + 17 v

Substitute back

= - 8 cos (arcsin (\frac{1}{8} (x - 8))) + 17 arcsin (\frac{1}{8} (x - 8))

Simplify

- 8 cos (arcsin (\frac{1}{8} (x - 8))) + 17 arcsin (\frac{1}{8} (x - 8)) : - - x^{2} + 16 x + 17 arcsin (\frac{1}{8} x - 1)

= - - x^{2} + 16 x + 17 arcsin (\frac{1}{8} x - 1)

Add a constant to the solution

= - - x^{2} + 16 x + 17 arcsin (\frac{1}{8} x - 1) + C

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What is the integral of (x+9)/(sqrt(64-(x-8)^2)) ?
The integral of (x+9)/(sqrt(64-(x-8)^2)) is -sqrt(-x^2+16x)+17arcsin(1/8 x-1)+C