What is the integral of (x+5)/(sqrt(16-(x-4)^2)) ?

The integral of (x+5)/(sqrt(16-(x-4)^2)) is -sqrt(-x^2+8x)+9arcsin(1/4 x-1)+C

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integral of (x+5)/(sqrt(16-(x-4)^2))

\int \frac{x + 5}{16 - ( x - 4 ) ^{2}} d x

- - x^{2} + 8 x + 9 arcsin (\frac{1}{4} x - 1) + C

Solution steps

\int \frac{x + 5}{16 - ( x - 4 ) ^{2}} d x

Apply u-substitution

: \int \frac{u + 9}{16 - u ^{2}} d u

= \int \frac{u + 9}{16 - u ^{2}} d u

Apply Trigonometric Substitution

: \int 4 sin (v) + 9 d v

= \int 4 sin (v) + 9 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 9 d v

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

\int 9 d v = 9 v

= - 4 cos (v) + 9 v

Substitute back

= - 4 cos (arcsin (\frac{1}{4} (x - 4))) + 9 arcsin (\frac{1}{4} (x - 4))

Simplify

- 4 cos (arcsin (\frac{1}{4} (x - 4))) + 9 arcsin (\frac{1}{4} (x - 4)) : - - x^{2} + 8 x + 9 arcsin (\frac{1}{4} x - 1)

= - - x^{2} + 8 x + 9 arcsin (\frac{1}{4} x - 1)

Add a constant to the solution

= - - x^{2} + 8 x + 9 arcsin (\frac{1}{4} x - 1) + C

f (x) = \frac{1}{5 x}

x \to 0 + lim (2 x + 8)

\frac{d}{d x} (tan (g (x)) (x))

x \to 4 lim (x^{2} + 3 x - 1)

x \to 0 lim (∣ 0 ∣)

What is the integral of (x+5)/(sqrt(16-(x-4)^2)) ?
The integral of (x+5)/(sqrt(16-(x-4)^2)) is -sqrt(-x^2+8x)+9arcsin(1/4 x-1)+C