What is the integral from 0 to 0.35 of 2/(x^2-4) ?

The integral from 0 to 0.35 of 2/(x^2-4) is -0.17682…

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integral from 0 to 0.35 of 2/(x^2-4)

\int_{0}^{0.35} \frac{2}{x ^{2} - 4} d x

- 0.17682 \dots

Solution steps

\int_{0}^{0.35} \frac{2}{x ^{2} - 4} d x

Take the constant out:

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

= 2 \cdot \int_{0}^{0.35} \frac{1}{x ^{2} - 4} d x

Apply Integral Substitution

= 2 \cdot \int_{0}^{0.175} \frac{1}{2 ( u ^{2} - 1 )} d u

Take the constant out:

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

= 2 \cdot \frac{1}{2} \cdot \int_{0}^{0.175} \frac{1}{u ^{2} - 1} d u

Factor

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

= 2 \cdot \frac{1}{2} \cdot \int_{0}^{0.175} \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

= 2 \cdot \frac{1}{2} (- \int_{0}^{0.175} \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}

= 2 \cdot \frac{1}{2} (- [\frac{ln ∣ u + 1 ∣}{2} - \frac{ln ∣ u - 1 ∣}{2}]_{0}^{0.175})

Simplify

2 \cdot \frac{1}{2} (- [\frac{ln ∣ u + 1 ∣}{2} - \frac{ln ∣ u - 1 ∣}{2}]_{0}^{0.175}) : - [0.5 ln ∣ u + 1 ∣ - 0.5 ln ∣ u - 1 ∣]_{0}^{0.175}

= - [0.5 ln ∣ u + 1 ∣ - 0.5 ln ∣ u - 1 ∣]_{0}^{0.175}

Compute the boundaries:

0.17682 \dots

= - 0.17682 \dots

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\int_{0}^{π} sec (2 x) d x

\int_{0}^{2} x^{2} + 4 d x

\int_{0}^{3} \frac{x}{1 + x + 1} d x

What is the integral from 0 to 0.35 of 2/(x^2-4) ?
The integral from 0 to 0.35 of 2/(x^2-4) is -0.17682…