What is the integral from 0 to 1 of (40)/(x^2-100) ?

The integral from 0 to 1 of (40)/(x^2-100) is -2(ln(11/10)-ln(9/10))

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integral from 0 to 1 of (40)/(x^2-100)

\int_{0}^{1} \frac{40}{x ^{2} - 100} d x

- 2 (ln (\frac{11}{10}) - ln (\frac{9}{10}))

Decimal Notation

- 0.40134 \dots

Solution steps

\int_{0}^{1} \frac{40}{x ^{2} - 100} d x

Take the constant out:

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

= 40 \cdot \int_{0}^{1} \frac{1}{x ^{2} - 100} d x

Apply Integral Substitution

= 40 \cdot \int_{0}^{\frac{1}{10}} \frac{1}{10 ( u ^{2} - 1 )} d u

Take the constant out:

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

= 40 \cdot \frac{1}{10} \cdot \int_{0}^{\frac{1}{10}} \frac{1}{u ^{2} - 1} d u

Factor

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

= 40 \cdot \frac{1}{10} \cdot \int_{0}^{\frac{1}{10}} \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

= 40 \cdot \frac{1}{10} (- \int_{0}^{\frac{1}{10}} \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}

= 40 \cdot \frac{1}{10} (- [\frac{ln ∣ u + 1 ∣}{2} - \frac{ln ∣ u - 1 ∣}{2}]_{0}^{\frac{1}{10}})

Simplify

40 \cdot \frac{1}{10} (- [\frac{ln ∣ u + 1 ∣}{2} - \frac{ln ∣ u - 1 ∣}{2}]_{0}^{\frac{1}{10}}) : - 4 [\frac{1}{2} (ln ∣ u + 1 ∣ - ln ∣ u - 1 ∣)]_{0}^{\frac{1}{10}}

= - 4 [\frac{1}{2} (ln ∣ u + 1 ∣ - ln ∣ u - 1 ∣)]_{0}^{\frac{1}{10}}

Compute the boundaries:

\frac{ln ( \frac{11}{10} ) - ln ( \frac{9}{10} )}{2}

= - 4 \cdot \frac{ln ( \frac{11}{10} ) - ln ( \frac{9}{10} )}{2}

Simplify

= - 2 (ln (\frac{11}{10}) - ln (\frac{9}{10}))

2^{x}, (0, 2)

x \to 8 lim (x^{2} - 9)

\int (x^{2} - 1)^{2} d x

x^{19} e^{x}

\frac{\partial}{\partial x} (x^{2} - 1)

What is the integral from 0 to 1 of (40)/(x^2-100) ?
The integral from 0 to 1 of (40)/(x^2-100) is -2(ln(11/10)-ln(9/10))