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

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

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

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

- \frac{1}{3} ln tan \frac{arctan ( \frac{1}{3} ( 1 - x ) )}{2} + C

Solution steps

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

Apply u-substitution

: \int - \frac{1}{u u ^{2} + 3} d u

= \int - \frac{1}{u u ^{2} + 3} d u

Take the constant out:

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

= - \int \frac{1}{u u ^{2} + 3} d u

Apply Trigonometric Substitution

: \int \frac{sec ( v )}{3 tan ( v )} d v

= - \int \frac{sec ( v )}{3 tan ( v )} d v

Take the constant out:

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

= - \frac{1}{3} \cdot \int \frac{sec ( v )}{tan ( v )} d v

Apply u-substitution

: \int \frac{1}{w} d w

= - \frac{1}{3} \cdot \int \frac{1}{w} d w

Use the common integral:

\int \frac{1}{w} d w = ln (∣ w ∣)

= - \frac{1}{3} ln ∣ w ∣

Substitute back

= - \frac{1}{3} ln tan \frac{arctan ( \frac{1}{3} ( 1 - x ) )}{2}

Add a constant to the solution

= - \frac{1}{3} ln tan \frac{arctan ( \frac{1}{3} ( 1 - x ) )}{2} + C

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