What is the integral of 6/((2-x)sqrt(x^2-4x+3)) ?

The integral of 6/((2-x)sqrt(x^2-4x+3)) is -6arctan(sqrt(x^2-4x+3))+C

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integral of 6/((2-x)sqrt(x^2-4x+3))

\int \frac{6}{( 2 - x ) x ^{2} - 4 x + 3} d x

- 6 arctan (x^{2} - 4 x + 3) + C

Solution steps

\int \frac{6}{( 2 - x ) x ^{2} - 4 x + 3} d x

Take the constant out:

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

= 6 \cdot \int \frac{1}{( 2 - x ) x ^{2} - 4 x + 3} d x

Apply u-substitution

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

= 6 \cdot \int - \frac{1}{u u ^{2} - 1} d u

Take the constant out:

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

= 6 (- \int \frac{1}{u u ^{2} - 1} d u)

Apply u-substitution

: \int \frac{1}{v ^{2} + 1} d v

= 6 (- \int \frac{1}{v ^{2} + 1} d v)

Use the common integral:

\int \frac{1}{v ^{2} + 1} d v = arctan (v)

= 6 (- arctan (v))

Substitute back

= 6 (- arctan ((2 - x)^{2} - 1))

Simplify

6 (- arctan ((2 - x)^{2} - 1)) : - 6 arctan (x^{2} - 4 x + 3)

= - 6 arctan (x^{2} - 4 x + 3)

Add a constant to the solution

= - 6 arctan (x^{2} - 4 x + 3) + C

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What is the integral of 6/((2-x)sqrt(x^2-4x+3)) ?
The integral of 6/((2-x)sqrt(x^2-4x+3)) is -6arctan(sqrt(x^2-4x+3))+C