What is the integral of (x+6)/(sqrt(25-(x-5)^2)) ?

The integral of (x+6)/(sqrt(25-(x-5)^2)) is -sqrt(-x^2+10x)+11arcsin(1/5 x-1)+C

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integral of (x+6)/(sqrt(25-(x-5)^2))

\int \frac{x + 6}{25 - ( x - 5 ) ^{2}} d x

- - x^{2} + 10 x + 11 arcsin (\frac{1}{5} x - 1) + C

Solution steps

\int \frac{x + 6}{25 - ( x - 5 ) ^{2}} d x

Apply u-substitution

: \int \frac{u + 11}{25 - u ^{2}} d u

= \int \frac{u + 11}{25 - u ^{2}} d u

Apply Trigonometric Substitution

: \int 5 sin (v) + 11 d v

= \int 5 sin (v) + 11 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 5 sin (v) d v + \int 11 d v

\int 5 sin (v) d v = - 5 cos (v)

\int 11 d v = 11 v

= - 5 cos (v) + 11 v

Substitute back

= - 5 cos (arcsin (\frac{1}{5} (x - 5))) + 11 arcsin (\frac{1}{5} (x - 5))

Simplify

- 5 cos (arcsin (\frac{1}{5} (x - 5))) + 11 arcsin (\frac{1}{5} (x - 5)) : - - x^{2} + 10 x + 11 arcsin (\frac{1}{5} x - 1)

= - - x^{2} + 10 x + 11 arcsin (\frac{1}{5} x - 1)

Add a constant to the solution

= - - x^{2} + 10 x + 11 arcsin (\frac{1}{5} x - 1) + C

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What is the integral of (x+6)/(sqrt(25-(x-5)^2)) ?
The integral of (x+6)/(sqrt(25-(x-5)^2)) is -sqrt(-x^2+10x)+11arcsin(1/5 x-1)+C