What is the integral from 1 to 2 of 8/(x^2-2x+2) ?

The integral from 1 to 2 of 8/(x^2-2x+2) is 2pi

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integral from 1 to 2 of 8/(x^2-2x+2)

\int_{1}^{2} \frac{8}{x ^{2} - 2 x + 2} d x

2 π

Decimal Notation

6.28318 \dots

Solution steps

\int_{1}^{2} \frac{8}{x ^{2} - 2 x + 2} d x

Take the constant out:

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

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

Complete the square

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

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

Apply u-substitution

: \int_{0}^{1} \frac{1}{u ^{2} + 1} d u

= 8 \cdot \int_{0}^{1} \frac{1}{u ^{2} + 1} d u

Use the common integral:

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

= 8 [arctan (u)]_{0}^{1}

Compute the boundaries:

\frac{π}{4}

= 8 \cdot \frac{π}{4}

Simplify

= 2 π

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

\int \frac{x ^{3}}{9 - x ^{2}} d x

\frac{\partial}{\partial y} (4 x^{2} + y^{2})

\frac{\partial}{\partial y} (8 - 2 x^{2} + y^{2})

f (x) = x^{2} (x^{2} + 1)

What is the integral from 1 to 2 of 8/(x^2-2x+2) ?
The integral from 1 to 2 of 8/(x^2-2x+2) is 2pi