What is the integral from 0 to 0.6 of 1/((1-x)(3-x)) ?

The integral from 0 to 0.6 of 1/((1-x)(3-x)) is 0.34657…

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integral from 0 to 0.6 of 1/((1-x)(3-x))

\int_{0}^{0.6} \frac{1}{( 1 - x ) ( 3 - x )} d x

0.34657 \dots

Solution steps

\int_{0}^{0.6} \frac{1}{( 1 - x ) ( 3 - x )} d x

Complete the square

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

= \int_{0}^{0.6} \frac{1}{( x - 2 ) ^{2} - 1} d x

Apply u-substitution

: \int_{- 2}^{- 1.4} \frac{1}{u ^{2} - 1} d u

= \int_{- 2}^{- 1.4} \frac{1}{u ^{2} - 1} d u

Factor

u^{2} - 1 : - (- u^{2} + 1)

= \int_{- 2}^{- 1.4} \frac{1}{- ( - u ^{2} + 1 )} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= - \int_{- 2}^{- 1.4} \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}]_{- 2}^{- 1.4}

Simplify

- [\frac{ln ∣ u + 1 ∣}{2} - \frac{ln ∣ u - 1 ∣}{2}]_{- 2}^{- 1.4} : - [\frac{1}{2} (ln ∣ u + 1 ∣ - ln ∣ u - 1 ∣)]_{- 2}^{- 1.4}

= - [\frac{1}{2} (ln ∣ u + 1 ∣ - ln ∣ u - 1 ∣)]_{- 2}^{- 1.4}

Compute the boundaries:

- 0.34657 \dots

= - (- 0.34657 \dots)

Simplify

= 0.34657 \dots

\int \frac{1}{t ^{2} 2 - t ^{2}} d t

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What is the integral from 0 to 0.6 of 1/((1-x)(3-x)) ?
The integral from 0 to 0.6 of 1/((1-x)(3-x)) is 0.34657…