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

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

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

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

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

Solution steps

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

Complete the square

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

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

Apply u-substitution

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

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

Apply Integral Substitution

= \int \frac{1}{3 ( v ^{2} + 1 )} 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{1}{v ^{2} + 1} d v

Use the common integral:

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

= \frac{1}{3} arctan (v)

Substitute back

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

Add a constant to the solution

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

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