What is the integral of x/(sqrt(9-(x-1)^2)) ?

The integral of x/(sqrt(9-(x-1)^2)) is -sqrt(-x^2+2x+8)+arcsin(1/3 (x-1))+C

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integral of x/(sqrt(9-(x-1)^2))

\int \frac{x}{9 - ( x - 1 ) ^{2}} d x

- - x^{2} + 2 x + 8 + arcsin (\frac{1}{3} (x - 1)) + C

Solution steps

\int \frac{x}{9 - ( x - 1 ) ^{2}} d x

Apply u-substitution

: \int \frac{u + 1}{9 - u ^{2}} d u

= \int \frac{u + 1}{9 - u ^{2}} d u

Expand

\frac{u + 1}{9 - u ^{2}} : \frac{u}{9 - u ^{2}} + \frac{1}{9 - u ^{2}}

= \int \frac{u}{9 - u ^{2}} + \frac{1}{9 - u ^{2}} d u

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int \frac{u}{9 - u ^{2}} d u + \int \frac{1}{9 - u ^{2}} d u

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

\int \frac{1}{9 - u ^{2}} d u = arcsin (\frac{1}{3} u)

= - 9 - u^{2} + arcsin (\frac{1}{3} u)

Substitute back

u = x - 1

= - 9 - (x - 1)^{2} + arcsin (\frac{1}{3} (x - 1))

Expand

9 - (x - 1)^{2} : - x^{2} + 2 x + 8

= - - x^{2} + 2 x + 8 + arcsin (\frac{1}{3} (x - 1))

Add a constant to the solution

= - - x^{2} + 2 x + 8 + arcsin (\frac{1}{3} (x - 1)) + C

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What is the integral of x/(sqrt(9-(x-1)^2)) ?
The integral of x/(sqrt(9-(x-1)^2)) is -sqrt(-x^2+2x+8)+arcsin(1/3 (x-1))+C