What is the integral from 1 to 3 of 1/(x^2-4x+5) ?

The integral from 1 to 3 of 1/(x^2-4x+5) is pi/2

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integral from 1 to 3 of 1/(x^2-4x+5)

\int_{1}^{3} \frac{1}{x ^{2} - 4 x + 5} d x

\frac{π}{2}

Decimal Notation

1.57079 \dots

Solution steps

\int_{1}^{3} \frac{1}{x ^{2} - 4 x + 5} d x

Complete the square

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

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

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

\int_{- 2}^{3} 1 - x^{2} d x

What is the integral from 1 to 3 of 1/(x^2-4x+5) ?
The integral from 1 to 3 of 1/(x^2-4x+5) is pi/2