What is the integral of 5/(4x^2-49) ?

The integral of 5/(4x^2-49) is -5/28 (ln| 2/7 x+1|-ln| 2/7 x-1|)+C

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integral of 5/(4x^2-49)

\int \frac{5}{4 x ^{2} - 49} d x

- \frac{5}{28} (ln \frac{2}{7} x + 1 - ln \frac{2}{7} x - 1) + C

Solution steps

\int \frac{5}{4 x ^{2} - 49} d x

Take the constant out:

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

= 5 \cdot \int \frac{1}{4 x ^{2} - 49} d x

Apply Integral Substitution

= 5 \cdot \int \frac{1}{14 ( u ^{2} - 1 )} d u

Take the constant out:

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

= 5 \cdot \frac{1}{14} \cdot \int \frac{1}{u ^{2} - 1} d u

Factor

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

= 5 \cdot \frac{1}{14} \cdot \int \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

= 5 \cdot \frac{1}{14} (- \int \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}

= 5 \cdot \frac{1}{14} (- (\frac{ln ∣ u + 1 ∣}{2} - \frac{ln ∣ u - 1 ∣}{2}))

Substitute back

u = \frac{2}{7} x

= 5 \cdot \frac{1}{14} (- (\frac{ln \frac{2}{7} x + 1}{2} - \frac{ln \frac{2}{7} x - 1}{2}))

Simplify

5 \cdot \frac{1}{14} (- (\frac{ln \frac{2}{7} x + 1}{2} - \frac{ln \frac{2}{7} x - 1}{2})) : - \frac{5}{28} (ln \frac{2}{7} x + 1 - ln \frac{2}{7} x - 1)

= - \frac{5}{28} (ln \frac{2}{7} x + 1 - ln \frac{2}{7} x - 1)

Add a constant to the solution

= - \frac{5}{28} (ln \frac{2}{7} x + 1 - ln \frac{2}{7} x - 1) + C

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What is the integral of 5/(4x^2-49) ?
The integral of 5/(4x^2-49) is -5/28 (ln| 2/7 x+1|-ln| 2/7 x-1|)+C