What is the integral of 1/(sqrt(-2x^2-12x+32)) ?

The integral of 1/(sqrt(-2x^2-12x+32)) is 1/(sqrt(2))arcsin((x+3)/5)+C

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integral of 1/(sqrt(-2x^2-12x+32))

\int \frac{1}{- 2 x ^{2} - 12 x + 32} d x

\frac{1}{2} arcsin (\frac{x + 3}{5}) + C

Solution steps

\int \frac{1}{- 2 x ^{2} - 12 x + 32} d x

Complete the square

- 2 x^{2} - 12 x + 32 : - 2 (x + 3)^{2} + 50

= \int \frac{1}{- 2 ( x + 3 ) ^{2} + 50} d x

Apply u-substitution

: \int \frac{1}{- 2 u ^{2} + 50} d u

= \int \frac{1}{- 2 u ^{2} + 50} 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}{5 2} (x + 3))

Simplify

\frac{1}{2} \cdot 1 \cdot arcsin (\frac{2}{5 2} (x + 3)) : \frac{1}{2} arcsin (\frac{x + 3}{5})

= \frac{1}{2} arcsin (\frac{x + 3}{5})

Add a constant to the solution

= \frac{1}{2} arcsin (\frac{x + 3}{5}) + C

3 t \frac{d y}{d t} + y = t^{4}

ln (\frac{e ^{x}}{1 + e ^{x}})

ln (\frac{a ^{2} - z ^{2}}{a ^{2} + z ^{2}})

x \to 5 lim (\frac{x - 5}{x - 5})

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

What is the integral of 1/(sqrt(-2x^2-12x+32)) ?
The integral of 1/(sqrt(-2x^2-12x+32)) is 1/(sqrt(2))arcsin((x+3)/5)+C