What is the integral of 1/(sqrt(-2x^2+8x+4)) ?

The integral of 1/(sqrt(-2x^2+8x+4)) is 1/(sqrt(2))arcsin(1/(sqrt(6))(x-2))+C

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integral of 1/(sqrt(-2x^2+8x+4))

\int \frac{1}{- 2 x ^{2} + 8 x + 4} d x

\frac{1}{2} arcsin (\frac{1}{6} (x - 2)) + C

Solution steps

\int \frac{1}{- 2 x ^{2} + 8 x + 4} d x

Complete the square

- 2 x^{2} + 8 x + 4 : - 2 (x - 2)^{2} + 12

= \int \frac{1}{- 2 ( x - 2 ) ^{2} + 12} d x

Apply u-substitution

: \int \frac{1}{- 2 u ^{2} + 12} d u

= \int \frac{1}{- 2 u ^{2} + 12} 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 3} (x - 2))

Simplify

\frac{1}{2} \cdot 1 \cdot arcsin (\frac{2}{2 3} (x - 2)) : \frac{1}{2} arcsin (\frac{1}{6} (x - 2))

= \frac{1}{2} arcsin (\frac{1}{6} (x - 2))

Add a constant to the solution

= \frac{1}{2} arcsin (\frac{1}{6} (x - 2)) + C

\int_{0}^{3} (2 x - x^{2})^{3} (1 - x) d x

\frac{\partial}{\partial x} (z^{2} cos (y x))

y^{^{'}} - \frac{6}{x} y = \frac{( y ^{3} )}{x ^{4}}

\int_{4}^{9} x x d x

\int \frac{1}{x ^{2} x ^{2} - 196} d x

What is the integral of 1/(sqrt(-2x^2+8x+4)) ?
The integral of 1/(sqrt(-2x^2+8x+4)) is 1/(sqrt(2))arcsin(1/(sqrt(6))(x-2))+C