What is the integral from 0 to 1 of 1/(x^2-4x+5) ?

The integral from 0 to 1 of 1/(x^2-4x+5) is -pi/4+arctan(2)

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integral from 0 to 1 of 1/(x^2-4x+5)

\int_{0}^{1} \frac{1}{x ^{2} - 4 x + 5} d x

- \frac{π}{4} + arctan (2)

Decimal

0.32175 \dots

Solution steps

\int_{0}^{1} \frac{1}{x ^{2} - 4 x + 5} d x

Complete the square

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

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

Apply u-substitution

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

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

Use the common integral:

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

= [arctan (u)]_{- 2}^{- 1}

Compute the boundaries:

- \frac{π}{4} + arctan (2)

= - \frac{π}{4} + arctan (2)

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What is the integral from 0 to 1 of 1/(x^2-4x+5) ?
The integral from 0 to 1 of 1/(x^2-4x+5) is -pi/4+arctan(2)