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

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

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

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

arctan (x - 1) + C

Solution steps

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

Complete the square

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

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

Apply u-substitution

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

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

Use the common integral:

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

= arctan (u)

Substitute back

u = x - 1

= arctan (x - 1)

Add a constant to the solution

= arctan (x - 1) + C

\int cos (x) 9 - sin^{2} (x) d x

x \to 2 lim (4 x - 1)

7 x y^{^{'}} - 2 y = \frac{- x ^{2}}{y ^{6}}

\int cos (x) + 1 d x

f (x) = \frac{x ^{2}}{2}, at x = 5

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