What is the integral from 0 to 1 of (28)/(x^2-49) ?

The integral from 0 to 1 of (28)/(x^2-49) is -2ln(4/3)

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integral from 0 to 1 of (28)/(x^2-49)

\int_{0}^{1} \frac{28}{x ^{2} - 49} d x

- 2 ln (\frac{4}{3})

Solution steps

\int_{0}^{1} \frac{28}{x ^{2} - 49} d x

Take the constant out:

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

= 28 \cdot \int_{0}^{1} \frac{1}{x ^{2} - 49} d x

Apply Integral Substitution

= 28 \cdot \int_{0}^{\frac{1}{7}} \frac{1}{7 ( u ^{2} - 1 )} d u

Take the constant out:

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

= 28 \cdot \frac{1}{7} \cdot \int_{0}^{\frac{1}{7}} \frac{1}{u ^{2} - 1} d u

Factor

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

= 28 \cdot \frac{1}{7} \cdot \int_{0}^{\frac{1}{7}} \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

= 28 \cdot \frac{1}{7} (- \int_{0}^{\frac{1}{7}} \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}

= 28 \cdot \frac{1}{7} (- [\frac{ln ∣ u + 1 ∣}{2} - \frac{ln ∣ u - 1 ∣}{2}]_{0}^{\frac{1}{7}})

Simplify

28 \cdot \frac{1}{7} (- [\frac{ln ∣ u + 1 ∣}{2} - \frac{ln ∣ u - 1 ∣}{2}]_{0}^{\frac{1}{7}}) : - 4 [\frac{1}{2} (ln ∣ u + 1 ∣ - ln ∣ u - 1 ∣)]_{0}^{\frac{1}{7}}

= - 4 [\frac{1}{2} (ln ∣ u + 1 ∣ - ln ∣ u - 1 ∣)]_{0}^{\frac{1}{7}}

Compute the boundaries:

\frac{ln ( \frac{8}{7} ) - ln ( \frac{6}{7} )}{2}

= - 4 \cdot \frac{ln ( \frac{8}{7} ) - ln ( \frac{6}{7} )}{2}

Simplify

= - 2 ln (\frac{4}{3})

6 y^{^{''}} - 6 y^{^{'}} + 15 y = 0

\frac{d}{d x} (xL (x) - x)

\int \frac{7 x}{x ^{3} + 1} d x

y = (x + 7) (7 x + 2)

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

What is the integral from 0 to 1 of (28)/(x^2-49) ?
The integral from 0 to 1 of (28)/(x^2-49) is -2ln(4/3)