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

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

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

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

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

Solution steps

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

Complete the square

- x^{2} + \frac{x}{2} + \frac{1}{2} : - (x - \frac{1}{4})^{2} + \frac{9}{16}

= \int \frac{1}{- ( x - \frac{1}{4} ) ^{2} + \frac{9}{16}} d x

Apply u-substitution

: \int \frac{4}{- 16 u ^{2} + 9} d u

= \int \frac{4}{- 16 u ^{2} + 9} d u

Take the constant out:

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

= 4 \cdot \int \frac{1}{- 16 u ^{2} + 9} d u

Apply Trigonometric Substitution

: \int \frac{1}{4} d v

= 4 \cdot \int \frac{1}{4} d v

Take the constant out:

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

= 4 \cdot \frac{1}{4} \cdot \int 1 d v

Integral of a constant:

\int a d x = a x

= 4 \cdot \frac{1}{4} \cdot 1 \cdot v

Substitute back

= 4 \cdot \frac{1}{4} \cdot 1 \cdot arcsin (\frac{4}{3} (x - \frac{1}{4}))

Simplify

4 \cdot \frac{1}{4} \cdot 1 \cdot arcsin (\frac{4}{3} (x - \frac{1}{4})) : arcsin (\frac{4}{3} x - \frac{1}{3})

= arcsin (\frac{4}{3} x - \frac{1}{3})

Add a constant to the solution

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

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