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

The integral of 1/(sqrt(2x^2-4x+6)) is 1/(sqrt(2))ln|(x-1}{sqrt(2)}+sqrt(\frac{x^2-2x+3)/2)|+C

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

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

\frac{1}{2} ln \frac{x - 1}{2} + \frac{x ^{2} - 2 x + 3}{2} + C

Solution steps

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

Complete the square

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

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

Apply u-substitution

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

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

Apply Trigonometric Substitution

: \int \frac{sec ( v )}{2} d v

= \int \frac{sec ( v )}{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 sec (v) d v

Use the common integral:

\int sec (v) d v = ln ∣ tan (v) + sec (v) ∣

= \frac{1}{2} ln ∣ tan (v) + sec (v) ∣

Substitute back

= \frac{1}{2} ln tan (arctan (\frac{2}{2} (x - 1))) + sec (arctan (\frac{2}{2} (x - 1)))

Simplify

\frac{1}{2} ln tan (arctan (\frac{2}{2} (x - 1))) + sec (arctan (\frac{2}{2} (x - 1))) : \frac{1}{2} ln \frac{x - 1}{2} + \frac{x ^{2} - 2 x + 3}{2}

= \frac{1}{2} ln \frac{x - 1}{2} + \frac{x ^{2} - 2 x + 3}{2}

Add a constant to the solution

= \frac{1}{2} ln \frac{x - 1}{2} + \frac{x ^{2} - 2 x + 3}{2} + C

\int_{0}^{1} 2 π x (33 - 33 x) d x

x \to 0 lim (\frac{\frac{4}{x + 4} - 1}{x})

\int_{0}^{1} \frac{73}{x ^{5}} d x

x \to - 1 lim (\frac{in x}{x - 1})

\frac{d y}{d x} = 2 x y^{2} + 72 x

What is the integral of 1/(sqrt(2x^2-4x+6)) ?
The integral of 1/(sqrt(2x^2-4x+6)) is 1/(sqrt(2))ln|(x-1}{sqrt(2)}+sqrt(\frac{x^2-2x+3)/2)|+C