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

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

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integral of 1/(sqrt(-x^2-x+2))

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

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

Solution steps

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

Complete the square

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

= \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

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

Apply Trigonometric Substitution

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

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

Integral of a constant:

\int a d x = a x

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

Substitute back

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

Simplify

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

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

Add a constant to the solution

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

f (x) = e^{x^{3} + 10 x}

\int 10 x + 2 d x

\int_{\frac{1}{2}}^{1} (x^{- 3} - 5) d x

y^{^{''}} = b y

\int_{1}^{2} \frac{x ^{2} + 5}{3 x - x ^{2}} d x

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