What is the integral of x/(sqrt(25-(3-x)^2)) ?

The integral of x/(sqrt(25-(3-x)^2)) is -sqrt(-x^2+6x+16)+3arcsin(1/5 (x-3))+C

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integral of x/(sqrt(25-(3-x)^2))

\int \frac{x}{25 - ( 3 - x ) ^{2}} d x

- - x^{2} + 6 x + 16 + 3 arcsin (\frac{1}{5} (x - 3)) + C

Solution steps

\int \frac{x}{25 - ( 3 - x ) ^{2}} d x

Expand

25 - (3 - x)^{2} : - x^{2} + 6 x + 16

= \int \frac{x}{- x ^{2} + 6 x + 16} d x

Complete the square

- x^{2} + 6 x + 16 : - (x - 3)^{2} + 25

= \int \frac{x}{- ( x - 3 ) ^{2} + 25} d x

Apply u-substitution

: \int \frac{u + 3}{- u ^{2} + 25} d u

= \int \frac{u + 3}{- u ^{2} + 25} d u

Apply Trigonometric Substitution

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

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

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

\int 3 d v = 3 v

= - 5 cos (v) + 3 v

Substitute back

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

Simplify

- 5 cos (arcsin (\frac{1}{5} (x - 3))) + 3 arcsin (\frac{1}{5} (x - 3)) : - - x^{2} + 6 x + 16 + 3 arcsin (\frac{1}{5} (x - 3))

= - - x^{2} + 6 x + 16 + 3 arcsin (\frac{1}{5} (x - 3))

Add a constant to the solution

= - - x^{2} + 6 x + 16 + 3 arcsin (\frac{1}{5} (x - 3)) + C

x \to 2 lim (- x^{2} + 7 x - 2)

\int 8 x^{3} sin (1 - x^{4}) d x

\int (x^{1.3} + 7 x^{2.5}) d x

\int_{1}^{5} \frac{24 + x ^{2}}{x} d x

\frac{d}{d x} (\frac{ln ( x )}{x ^{8}})

What is the integral of x/(sqrt(25-(3-x)^2)) ?
The integral of x/(sqrt(25-(3-x)^2)) is -sqrt(-x^2+6x+16)+3arcsin(1/5 (x-3))+C