What is the integral of 1/(sqrt(4x^2-4x+3)) ?

The integral of 1/(sqrt(4x^2-4x+3)) is 1/2 ln|sqrt(2)(x-1/2)+sqrt(1+2(x^2-x+1/4))|+C

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integral of 1/(sqrt(4x^2-4x+3))

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

\frac{1}{2} ln 2 (x - \frac{1}{2}) + 1 + 2 (x^{2} - x + \frac{1}{4}) + C

Solution steps

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

Complete the square

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

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

Apply u-substitution

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

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

Simplify

\frac{1}{2} ln tan (arctan (\frac{2}{2} (x - \frac{1}{2}))) + sec (arctan (\frac{2}{2} (x - \frac{1}{2}))) : \frac{1}{2} ln 2 (x - \frac{1}{2}) + 1 + 2 (x^{2} - x + \frac{1}{4})

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

Add a constant to the solution

= \frac{1}{2} ln 2 (x - \frac{1}{2}) + 1 + 2 (x^{2} - x + \frac{1}{4}) + C

\frac{d}{d x} (x^{3} - \frac{1}{12 x})

\frac{5}{x ^{3}}, - 1

\frac{\partial}{\partial x} (- 2 y cos (6 x))

\frac{\partial}{\partial y} (\frac{1}{x ^{2}} + \frac{1}{y ^{2}})

\frac{\partial}{\partial u ( v )} (\frac{e ^{v}}{u ( v ) + v ^{6}})

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