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

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

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

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

2 arctan (2 x - 1) + C

Solution steps

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

Complete the square

x^{2} - 2 x + 1 : (x - \frac{1}{2})^{2} + \frac{1}{2}

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

Apply u-substitution

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

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

Take the constant out:

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

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

Apply Integral Substitution

= 2 \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

= 2 \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)

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

Substitute back

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

Simplify

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

= 2 arctan (2 x - 1)

Add a constant to the solution

= 2 arctan (2 x - 1) + C

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