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

The integral of 1/(sqrt(3+x-x^2)) is arcsin(1/(sqrt(13))(2x-1))+C

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integral of 1/(sqrt(3+x-x^2))

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

arcsin (\frac{1}{13} (2 x - 1)) + C

Solution steps

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

Complete the square

3 + x - x^{2} : - (x - \frac{1}{2})^{2} + \frac{13}{4}

= \int \frac{1}{- ( x - \frac{1}{2} ) ^{2} + \frac{13}{4}} d x

Apply u-substitution

: \int \frac{2}{- 4 u ^{2} + 13} d u

= \int \frac{2}{- 4 u ^{2} + 13} 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} + 13} 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}{13} (x - \frac{1}{2}))

Simplify

2 \cdot \frac{1}{2} \cdot 1 \cdot arcsin (\frac{2}{13} (x - \frac{1}{2})) : arcsin (\frac{1}{13} (2 x - 1))

= arcsin (\frac{1}{13} (2 x - 1))

Add a constant to the solution

= arcsin (\frac{1}{13} (2 x - 1)) + C

(x^{3} - 2 x + 1) (3 x^{3} + 2 x^{2} - 5 x)

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

\frac{( 3 y ^{2} - t ^{2} )}{y ^{5}} d y + \frac{t}{2 y ^{4}} d t = 0

n = 0 \sum \infty (\frac{3}{5})^{n}

x \to + 3 - lim (\frac{3}{x - 3})

What is the integral of 1/(sqrt(3+x-x^2)) ?
The integral of 1/(sqrt(3+x-x^2)) is arcsin(1/(sqrt(13))(2x-1))+C