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

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

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

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

\frac{ln ( 3 )}{2}

Decimal

0.54930 \dots

Solution steps

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

Complete the square

- x^{2} + 4 x - 3 : - (x - 2)^{2} + 1

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

Apply u-substitution

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

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

Use the common integral:

\int \frac{1}{- u ^{2} + 1} d u = \frac{ln ∣ u + 1 ∣}{2} - \frac{ln ∣ u - 1 ∣}{2}

= [\frac{ln ∣ u + 1 ∣}{2} - \frac{ln ∣ u - 1 ∣}{2}]_{- \frac{1}{2}}^{0}

Simplify

[\frac{ln ∣ u + 1 ∣}{2} - \frac{ln ∣ u - 1 ∣}{2}]_{- \frac{1}{2}}^{0} : [\frac{1}{2} (ln ∣ u + 1 ∣ - ln ∣ u - 1 ∣)]_{- \frac{1}{2}}^{0}

= [\frac{1}{2} (ln ∣ u + 1 ∣ - ln ∣ u - 1 ∣)]_{- \frac{1}{2}}^{0}

Compute the boundaries:

- \frac{- ln ( 2 ) - ln ( \frac{3}{2} )}{2}

= - \frac{- ln ( 2 ) - ln ( \frac{3}{2} )}{2}

Simplify

= \frac{ln ( 3 )}{2}

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