What is the integral of (x+7)/(sqrt(36-(x-6)^2)) ?

The integral of (x+7)/(sqrt(36-(x-6)^2)) is -sqrt(-x^2+12x)+13arcsin(1/6 x-1)+C

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integral of (x+7)/(sqrt(36-(x-6)^2))

\int \frac{x + 7}{36 - ( x - 6 ) ^{2}} d x

- - x^{2} + 12 x + 13 arcsin (\frac{1}{6} x - 1) + C

Solution steps

\int \frac{x + 7}{36 - ( x - 6 ) ^{2}} d x

Apply u-substitution

: \int \frac{u + 13}{36 - u ^{2}} d u

= \int \frac{u + 13}{36 - u ^{2}} d u

Apply Trigonometric Substitution

: \int 6 sin (v) + 13 d v

= \int 6 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 6 sin (v) d v + \int 13 d v

\int 6 sin (v) d v = - 6 cos (v)

\int 13 d v = 13 v

= - 6 cos (v) + 13 v

Substitute back

= - 6 cos (arcsin (\frac{1}{6} (x - 6))) + 13 arcsin (\frac{1}{6} (x - 6))

Simplify

- 6 cos (arcsin (\frac{1}{6} (x - 6))) + 13 arcsin (\frac{1}{6} (x - 6)) : - - x^{2} + 12 x + 13 arcsin (\frac{1}{6} x - 1)

= - - x^{2} + 12 x + 13 arcsin (\frac{1}{6} x - 1)

Add a constant to the solution

= - - x^{2} + 12 x + 13 arcsin (\frac{1}{6} x - 1) + C

\int \frac{x ^{2}}{( x ^{2} + 100 ) ^{3}} d x

\int \frac{1}{1 - 5 y ^{2}} d y

\int \frac{x ^{2}}{a ^{2} + x ^{2}} d x

\int sin (x - 1) d x

What is the integral of (x+7)/(sqrt(36-(x-6)^2)) ?
The integral of (x+7)/(sqrt(36-(x-6)^2)) is -sqrt(-x^2+12x)+13arcsin(1/6 x-1)+C