What is the integral of 1/((x-3)sqrt(x^2-6x+8)) ?

The integral of 1/((x-3)sqrt(x^2-6x+8)) is arctan(sqrt(x^2-6x+8))+C

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integral of 1/((x-3)sqrt(x^2-6x+8))

\int \frac{1}{( x - 3 ) x ^{2} - 6 x + 8} d x

arctan (x^{2} - 6 x + 8) + C

Solution steps

\int \frac{1}{( x - 3 ) x ^{2} - 6 x + 8} d x

Apply u-substitution

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

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

Apply u-substitution

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

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

Use the common integral:

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

= arctan (v)

Substitute back

= arctan ((x - 3)^{2} - 1)

Expand

(x - 3)^{2} - 1 : x^{2} - 6 x + 8

= arctan (x^{2} - 6 x + 8)

Add a constant to the solution

= arctan (x^{2} - 6 x + 8) + C

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\int 2 - 3 x d x

What is the integral of 1/((x-3)sqrt(x^2-6x+8)) ?
The integral of 1/((x-3)sqrt(x^2-6x+8)) is arctan(sqrt(x^2-6x+8))+C