What is the integral of 1/(sqrt(6+4x-2x^2)) ?

The integral of 1/(sqrt(6+4x-2x^2)) is 1/(sqrt(2))arcsin((x-1)/2)+C

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integral of 1/(sqrt(6+4x-2x^2))

\int \frac{1}{6 + 4 x - 2 x ^{2}} d x

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

Solution steps

\int \frac{1}{6 + 4 x - 2 x ^{2}} d x

Complete the square

6 + 4 x - 2 x^{2} : - 2 (x - 1)^{2} + 8

= \int \frac{1}{- 2 ( x - 1 ) ^{2} + 8} d x

Apply u-substitution

: \int \frac{1}{- 2 u ^{2} + 8} d u

= \int \frac{1}{- 2 u ^{2} + 8} d u

Apply Trigonometric Substitution

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

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

Take the constant out:

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

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

Integral of a constant:

\int a d x = a x

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

Substitute back

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

Simplify

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

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

Add a constant to the solution

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

ln (81 sin^{2} (x))

\int \frac{200}{20 - 3 x} d x

x \to 1 lim (x^{3} + 1)

x \to \infty lim (5 \cdot x)

\int \frac{1}{( 1 + x ) x} d x

What is the integral of 1/(sqrt(6+4x-2x^2)) ?
The integral of 1/(sqrt(6+4x-2x^2)) is 1/(sqrt(2))arcsin((x-1)/2)+C