What is the integral of (2x)/(sqrt(5+4x-x^2)) ?

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

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integral of (2x)/(sqrt(5+4x-x^2))

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

2 (- - x^{2} + 4 x + 5 + 2 arcsin (\frac{1}{3} (x - 2))) + C

Solution steps

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

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

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

Complete the square

5 + 4 x - x^{2} : - (x - 2)^{2} + 9

= 2 \cdot \int \frac{x}{- ( x - 2 ) ^{2} + 9} d x

Apply u-substitution

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

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

Apply Trigonometric Substitution

: \int 3 sin (v) + 2 d v

= 2 \cdot \int 3 sin (v) + 2 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

= 2 (\int 3 sin (v) d v + \int 2 d v)

\int 3 sin (v) d v = - 3 cos (v)

\int 2 d v = 2 v

= 2 (- 3 cos (v) + 2 v)

Substitute back

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

Simplify

2 (- 3 cos (arcsin (\frac{1}{3} (x - 2))) + 2 arcsin (\frac{1}{3} (x - 2))) : 2 (- - x^{2} + 4 x + 5 + 2 arcsin (\frac{1}{3} (x - 2)))

= 2 (- - x^{2} + 4 x + 5 + 2 arcsin (\frac{1}{3} (x - 2)))

Add a constant to the solution

= 2 (- - x^{2} + 4 x + 5 + 2 arcsin (\frac{1}{3} (x - 2))) + C

\int x^{3} x^{2} + 23 d x

\int 0.05 x + 2 d x

\int \frac{4}{( x + 4 ) ^{5}} d x

\int \frac{2 x}{x ^{2} + 2 x + 5} d x

\int \frac{t ^{2}}{t ^{2} + 1} d t

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