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

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

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

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

arcsin (2 x - 3) + C

Solution steps

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

Complete the square

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

= \int \frac{1}{- ( x - \frac{3}{2} ) ^{2} + \frac{1}{4}} d x

Apply u-substitution

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

= \int \frac{2}{- 4 u ^{2} + 1} 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} + 1} 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 (2 (x - \frac{3}{2}))

Simplify

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

= arcsin (2 x - 3)

Add a constant to the solution

= arcsin (2 x - 3) + C

f (x) = \frac{3 - x e ^{x}}{x + e ^{x}}

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

\int_{0}^{π} (sin (x)) d x

\int (x 7 - x^{2}) d x

x \to 4 lim (x^{3} - c x)

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