What is the integral from 1 to 3 of 1/(x^2-2x+5) ?

The integral from 1 to 3 of 1/(x^2-2x+5) is pi/8

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integral from 1 to 3 of 1/(x^2-2x+5)

\int_{1}^{3} \frac{1}{x ^{2} - 2 x + 5} d x

\frac{π}{8}

Decimal Notation

0.39269 \dots

Solution steps

\int_{1}^{3} \frac{1}{x ^{2} - 2 x + 5} d x

Complete the square

x^{2} - 2 x + 5 : (x - 1)^{2} + 4

= \int_{1}^{3} \frac{1}{( x - 1 ) ^{2} + 4} d x

Apply u-substitution

: \int_{0}^{2} \frac{1}{u ^{2} + 4} d u

= \int_{0}^{2} \frac{1}{u ^{2} + 4} d u

Apply Integral Substitution

= \int_{0}^{1} \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

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

Use the common integral:

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

= \frac{1}{2} [arctan (v)]_{0}^{1}

Compute the boundaries:

\frac{π}{4}

= \frac{1}{2} \cdot \frac{π}{4}

Simplify

= \frac{π}{8}

\int e^{i 7 x} cos (4 x) d x

\int e^{2 x} x^{3} d x

\frac{( s - 1 )}{4 s ^{2} + s + 24}

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

y^{^{'}} = \frac{4 t ^{3} + 1}{2 y - 6}, y (1) = 2

What is the integral from 1 to 3 of 1/(x^2-2x+5) ?
The integral from 1 to 3 of 1/(x^2-2x+5) is pi/8