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

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

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

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

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

Solution steps

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

Take the constant out:

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

= 2 \cdot \int \frac{1}{- x ^{2} + 4 x} d x

Complete the square

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

= 2 \cdot \int \frac{1}{- ( x - 2 ) ^{2} + 4} d x

Apply u-substitution

: \int \frac{1}{- u ^{2} + 4} d u

= 2 \cdot \int \frac{1}{- u ^{2} + 4} d u

Apply Trigonometric Substitution

: \int 1 d v

= 2 \cdot \int 1 d v

Integral of a constant:

\int a d x = a x

= 2 \cdot 1 \cdot v

Substitute back

= 2 \cdot 1 \cdot arcsin (\frac{1}{2} (x - 2))

Simplify

2 \cdot 1 \cdot arcsin (\frac{1}{2} (x - 2)) : 2 arcsin (\frac{1}{2} x - 1)

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

Add a constant to the solution

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

\frac{d}{d θ} (4 + 4 sin (θ))

\frac{d}{d x} (x - 3 x^{\frac{1}{3}})

\frac{d}{d x} (cos (3 x) \cdot 3)

f (x) = \frac{x ^{3}}{3}

f (x) = x - e^{x}

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