What is the integral of (x+3)/(sqrt(4-(x-2)^2)) ?

The integral of (x+3)/(sqrt(4-(x-2)^2)) is -sqrt(-x^2+4x)+5arcsin(1/2 x-1)+C

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integral of (x+3)/(sqrt(4-(x-2)^2))

\int \frac{x + 3}{4 - ( x - 2 ) ^{2}} d x

- - x^{2} + 4 x + 5 arcsin (\frac{1}{2} x - 1) + C

Solution steps

\int \frac{x + 3}{4 - ( x - 2 ) ^{2}} d x

Apply u-substitution

: \int \frac{u + 5}{4 - u ^{2}} d u

= \int \frac{u + 5}{4 - u ^{2}} d u

Apply Trigonometric Substitution

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

= \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

= \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

= - 2 cos (v) + 5 v

Substitute back

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

Simplify

- 2 cos (arcsin (\frac{1}{2} (x - 2))) + 5 arcsin (\frac{1}{2} (x - 2)) : - - x^{2} + 4 x + 5 arcsin (\frac{1}{2} x - 1)

= - - x^{2} + 4 x + 5 arcsin (\frac{1}{2} x - 1)

Add a constant to the solution

= - - x^{2} + 4 x + 5 arcsin (\frac{1}{2} x - 1) + C

(2 x + 3) y^{^{'}} = y + 2 x + 3

\frac{1 d y}{x ^{2} d x} = y^{2} + 9

2^{90}

f (x) = 2 x^{2} - 7 x + 5

\frac{d}{d x} (\frac{e ^{4 x^{2} + 1}}{8})

What is the integral of (x+3)/(sqrt(4-(x-2)^2)) ?
The integral of (x+3)/(sqrt(4-(x-2)^2)) is -sqrt(-x^2+4x)+5arcsin(1/2 x-1)+C