What is the integral from-3 to-1 of 1/(x^2-6x+25) ?

The integral from-3 to-1 of 1/(x^2-6x+25) is 1/4 (-pi/4+arctan(3/2))

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integral from-3 to-1 of 1/(x^2-6x+25)

\int_{- 3}^{- 1} \frac{1}{x ^{2} - 6 x + 25} d x

\frac{1}{4} (- \frac{π}{4} + arctan (\frac{3}{2}))

Decimal

0.04934 \dots

Solution steps

\int_{- 3}^{- 1} \frac{1}{x ^{2} - 6 x + 25} d x

Complete the square

x^{2} - 6 x + 25 : (x - 3)^{2} + 16

= \int_{- 3}^{- 1} \frac{1}{( x - 3 ) ^{2} + 16} d x

Apply u-substitution

: \int_{- 6}^{- 4} \frac{1}{u ^{2} + 16} d u

= \int_{- 6}^{- 4} \frac{1}{u ^{2} + 16} d u

Apply Integral Substitution

= \int_{- \frac{3}{2}}^{- 1} \frac{1}{4 ( 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}{4} \cdot \int_{- \frac{3}{2}}^{- 1} \frac{1}{v ^{2} + 1} d v

Use the common integral:

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

= \frac{1}{4} [arctan (v)]_{- \frac{3}{2}}^{- 1}

Compute the boundaries:

- \frac{π}{4} + arctan (\frac{3}{2})

= \frac{1}{4} (- \frac{π}{4} + arctan (\frac{3}{2}))

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What is the integral from-3 to-1 of 1/(x^2-6x+25) ?
The integral from-3 to-1 of 1/(x^2-6x+25) is 1/4 (-pi/4+arctan(3/2))