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

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

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

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

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

Solution steps

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

Complete the square

2 + x - x^{2} : - (x - \frac{1}{2})^{2} + \frac{9}{4}

= 2 \cdot \int \frac{1}{- ( x - \frac{1}{2} ) ^{2} + \frac{9}{4}} d x

Apply u-substitution

: \int \frac{2}{- 4 u ^{2} + 9} d u

= 2 \cdot \int \frac{2}{- 4 u ^{2} + 9} d u

Take the constant out:

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

= 2 \cdot 2 \cdot \int \frac{1}{- 4 u ^{2} + 9} d u

Apply Trigonometric Substitution

: \int \frac{1}{2} d v

= 2 \cdot 2 \cdot \int \frac{1}{2} d v

Take the constant out:

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

= 2 \cdot 2 \cdot \frac{1}{2} \cdot \int 1 d v

Integral of a constant:

\int a d x = a x

= 2 \cdot 2 \cdot \frac{1}{2} \cdot 1 \cdot v

Substitute back

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

Simplify

2 \cdot 2 \cdot \frac{1}{2} \cdot 1 \cdot arcsin (\frac{2}{3} (x - \frac{1}{2})) : 2 arcsin (\frac{2}{3} x - \frac{1}{3})

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

Add a constant to the solution

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

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