What is the integral from-1 to 2 of 1/(x^2-4x+13) ?

The integral from-1 to 2 of 1/(x^2-4x+13) is pi/(12)

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integral from-1 to 2 of 1/(x^2-4x+13)

\int_{- 1}^{2} \frac{1}{x ^{2} - 4 x + 13} d x

\frac{π}{12}

Decimal Notation

0.26179 \dots

Solution steps

\int_{- 1}^{2} \frac{1}{x ^{2} - 4 x + 13} d x

Complete the square

x^{2} - 4 x + 13 : (x - 2)^{2} + 9

= \int_{- 1}^{2} \frac{1}{( x - 2 ) ^{2} + 9} d x

Apply u-substitution

: \int_{- 3}^{0} \frac{1}{u ^{2} + 9} d u

= \int_{- 3}^{0} \frac{1}{u ^{2} + 9} d u

Apply Integral Substitution

= \int_{- 1}^{0} \frac{1}{3 ( 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}{3} \cdot \int_{- 1}^{0} \frac{1}{v ^{2} + 1} d v

Use the common integral:

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

= \frac{1}{3} [arctan (v)]_{- 1}^{0}

Compute the boundaries:

\frac{π}{4}

= \frac{1}{3} \cdot \frac{π}{4}

Simplify

= \frac{π}{12}

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What is the integral from-1 to 2 of 1/(x^2-4x+13) ?
The integral from-1 to 2 of 1/(x^2-4x+13) is pi/(12)