What is the integral from 3 to 5 of 1/(x^2-4x+8) ?

The integral from 3 to 5 of 1/(x^2-4x+8) is 1/2 (arctan(3/2)-arctan(1/2))

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integral from 3 to 5 of 1/(x^2-4x+8)

\int_{3}^{5} \frac{1}{x ^{2} - 4 x + 8} d x

\frac{1}{2} (arctan (\frac{3}{2}) - arctan (\frac{1}{2}))

Decimal

0.25957 \dots

Solution steps

\int_{3}^{5} \frac{1}{x ^{2} - 4 x + 8} d x

Complete the square

x^{2} - 4 x + 8 : (x - 2)^{2} + 4

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

Apply u-substitution

: \int_{1}^{3} \frac{1}{u ^{2} + 4} d u

= \int_{1}^{3} \frac{1}{u ^{2} + 4} d u

Apply Integral Substitution

= \int_{\frac{1}{2}}^{\frac{3}{2}} \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_{\frac{1}{2}}^{\frac{3}{2}} \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)]_{\frac{1}{2}}^{\frac{3}{2}}

Compute the boundaries:

arctan (\frac{3}{2}) - arctan (\frac{1}{2})

= \frac{1}{2} (arctan (\frac{3}{2}) - arctan (\frac{1}{2}))

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What is the integral from 3 to 5 of 1/(x^2-4x+8) ?
The integral from 3 to 5 of 1/(x^2-4x+8) is 1/2 (arctan(3/2)-arctan(1/2))