What is the integral of (x+9)/(sqrt(36-(x-6)^2)) ?

The integral of (x+9)/(sqrt(36-(x-6)^2)) is -sqrt(-x^2+12x)+15arcsin(1/6 x-1)+C

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integral of (x+9)/(sqrt(36-(x-6)^2))

\int \frac{x + 9}{36 - ( x - 6 ) ^{2}} d x

- - x^{2} + 12 x + 15 arcsin (\frac{1}{6} x - 1) + C

Solution steps

\int \frac{x + 9}{36 - ( x - 6 ) ^{2}} d x

Apply u-substitution

: \int \frac{u + 15}{36 - u ^{2}} d u

= \int \frac{u + 15}{36 - u ^{2}} d u

Apply Trigonometric Substitution

: \int 3 (2 sin (v) + 5) d v

= \int 3 (2 sin (v) + 5) d v

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 3 \cdot \int 2 sin (v) + 5 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

= 3 (\int 2 sin (v) d v + \int 5 d v)

\int 2 sin (v) d v = - 2 cos (v)

\int 5 d v = 5 v

= 3 (- 2 cos (v) + 5 v)

Substitute back

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

Simplify

3 (- 2 cos (arcsin (\frac{1}{6} (x - 6))) + 5 arcsin (\frac{1}{6} (x - 6))) : - - x^{2} + 12 x + 15 arcsin (\frac{1}{6} x - 1)

= - - x^{2} + 12 x + 15 arcsin (\frac{1}{6} x - 1)

Add a constant to the solution

= - - x^{2} + 12 x + 15 arcsin (\frac{1}{6} x - 1) + C

n = 1 \sum \infty \frac{2 n}{5 n - 3}

e^{s i n (2 x)}

\int_{0}^{1} 2 (1 - x) d x

\int \frac{4 x}{4 x ^{2} + 1} d x

\frac{d}{d x} (\frac{2 x}{( x ^{2} - 3 ) ^{2}})

What is the integral of (x+9)/(sqrt(36-(x-6)^2)) ?
The integral of (x+9)/(sqrt(36-(x-6)^2)) is -sqrt(-x^2+12x)+15arcsin(1/6 x-1)+C