What is the integral from 0 to 1 of (32)/(x^2-64) ?

The integral from 0 to 1 of (32)/(x^2-64) is -2(ln(9/8)-ln(7/8))

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integral from 0 to 1 of (32)/(x^2-64)

\int_{0}^{1} \frac{32}{x ^{2} - 64} d x

- 2 (ln (\frac{9}{8}) - ln (\frac{7}{8}))

Decimal Notation

- 0.50262 \dots

Solution steps

\int_{0}^{1} \frac{32}{x ^{2} - 64} d x

Take the constant out:

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

= 32 \cdot \int_{0}^{1} \frac{1}{x ^{2} - 64} d x

Apply Integral Substitution

= 32 \cdot \int_{0}^{\frac{1}{8}} \frac{1}{8 ( u ^{2} - 1 )} d u

Take the constant out:

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

= 32 \cdot \frac{1}{8} \cdot \int_{0}^{\frac{1}{8}} \frac{1}{u ^{2} - 1} d u

Factor

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

= 32 \cdot \frac{1}{8} \cdot \int_{0}^{\frac{1}{8}} \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

= 32 \cdot \frac{1}{8} (- \int_{0}^{\frac{1}{8}} \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}

= 32 \cdot \frac{1}{8} (- [\frac{ln ∣ u + 1 ∣}{2} - \frac{ln ∣ u - 1 ∣}{2}]_{0}^{\frac{1}{8}})

Simplify

32 \cdot \frac{1}{8} (- [\frac{ln ∣ u + 1 ∣}{2} - \frac{ln ∣ u - 1 ∣}{2}]_{0}^{\frac{1}{8}}) : - 4 [\frac{1}{2} (ln ∣ u + 1 ∣ - ln ∣ u - 1 ∣)]_{0}^{\frac{1}{8}}

= - 4 [\frac{1}{2} (ln ∣ u + 1 ∣ - ln ∣ u - 1 ∣)]_{0}^{\frac{1}{8}}

Compute the boundaries:

\frac{ln ( \frac{9}{8} ) - ln ( \frac{7}{8} )}{2}

= - 4 \cdot \frac{ln ( \frac{9}{8} ) - ln ( \frac{7}{8} )}{2}

Simplify

= - 2 (ln (\frac{9}{8}) - ln (\frac{7}{8}))

\frac{1}{s ^{2} \cdot ( s + 1 )}

f (x) = \frac{7}{2} x^{\frac{2}{7}}

x \to 8 lim (\frac{x ^{2} - 64}{x ^{2} - 8 x})

\frac{\partial}{\partial x} (2 x + 1)

\int (\frac{8}{x ^{3}}) d x

What is the integral from 0 to 1 of (32)/(x^2-64) ?
The integral from 0 to 1 of (32)/(x^2-64) is -2(ln(9/8)-ln(7/8))