What is the integral from 0 to 2 of 1/(x^2-2x+2) ?

The integral from 0 to 2 of 1/(x^2-2x+2) is pi/2

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integral from 0 to 2 of 1/(x^2-2x+2)

\int_{0}^{2} \frac{1}{x ^{2} - 2 x + 2} d x

\frac{π}{2}

Decimal Notation

1.57079 \dots

Solution steps

\int_{0}^{2} \frac{1}{x ^{2} - 2 x + 2} d x

Complete the square

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

= \int_{0}^{2} \frac{1}{( x - 1 ) ^{2} + 1} d x

Apply u-substitution

: \int_{- 1}^{1} \frac{1}{u ^{2} + 1} d u

= \int_{- 1}^{1} \frac{1}{u ^{2} + 1} d u

Use the common integral:

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

= [arctan (u)]_{- 1}^{1}

Compute the boundaries:

\frac{π}{2}

= \frac{π}{2}

\int \frac{1}{a + b x ^{2}} d x

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

x \to 0 lim (\frac{0}{2 x ^{2}})

\frac{d y}{d t} = sin (t) - y, y (0) = - 1

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

What is the integral from 0 to 2 of 1/(x^2-2x+2) ?
The integral from 0 to 2 of 1/(x^2-2x+2) is pi/2