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

The integral of 8/(x^2-2x+3) is 4sqrt(2)arctan(1/(sqrt(2))(x-1))+C

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

\int \frac{8}{x ^{2} - 2 x + 3} d x

42 arctan (\frac{1}{2} (x - 1)) + C

Solution steps

\int \frac{8}{x ^{2} - 2 x + 3} d x

Take the constant out:

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

= 8 \cdot \int \frac{1}{x ^{2} - 2 x + 3} d x

Complete the square

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

= 8 \cdot \int \frac{1}{( x - 1 ) ^{2} + 2} d x

Apply u-substitution

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

= 8 \cdot \int \frac{1}{u ^{2} + 2} d u

Apply Integral Substitution

= 8 \cdot \int \frac{1}{2 ( v ^{2} + 1 )} d v

Take the constant out:

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

= 8 \cdot \frac{1}{2} \cdot \int \frac{1}{v ^{2} + 1} d v

Use the common integral:

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

= 8 \cdot \frac{1}{2} arctan (v)

Substitute back

= 8 \cdot \frac{1}{2} arctan (\frac{x - 1}{2})

Simplify

8 \cdot \frac{1}{2} arctan (\frac{x - 1}{2}) : 42 arctan (\frac{1}{2} (x - 1))

= 42 arctan (\frac{1}{2} (x - 1))

Add a constant to the solution

= 42 arctan (\frac{1}{2} (x - 1)) + C

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What is the integral of 8/(x^2-2x+3) ?
The integral of 8/(x^2-2x+3) is 4sqrt(2)arctan(1/(sqrt(2))(x-1))+C