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

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

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

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

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

Solution steps

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

Apply radical rule:

a = a^{\frac{1}{2}}

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

Apply exponent rule:

\frac{1}{a ^{b}} = a^{- b}

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

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

Complete the square

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

= \int (- (x - \frac{1}{2})^{2} + \frac{5}{4})^{- \frac{1}{2}} d x

Apply u-substitution

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

= \int \frac{2}{- 4 u ^{2} + 5} 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} + 5} 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}{5} (x - \frac{1}{2}))

Simplify

2 \cdot \frac{1}{2} \cdot 1 \cdot arcsin (\frac{2}{5} (x - \frac{1}{2})) : arcsin (\frac{1}{5} (2 x - 1))

= arcsin (\frac{1}{5} (2 x - 1))

Add a constant to the solution

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

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