What is the integral of 1/(16x^2-25) ?

The integral of 1/(16x^2-25) is -1/40 (ln| 4/5 x+1|-ln| 4/5 x-1|)+C

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integral of 1/(16x^2-25)

\int \frac{1}{16 x ^{2} - 25} d x

- \frac{1}{40} (ln \frac{4}{5} x + 1 - ln \frac{4}{5} x - 1) + C

Solution steps

\int \frac{1}{16 x ^{2} - 25} d x

Apply Integral Substitution

= \int \frac{1}{20 ( u ^{2} - 1 )} d u

Take the constant out:

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

= \frac{1}{20} \cdot \int \frac{1}{u ^{2} - 1} d u

Factor

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

= \frac{1}{20} \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

= \frac{1}{20} (- \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}

= \frac{1}{20} (- (\frac{ln ∣ u + 1 ∣}{2} - \frac{ln ∣ u - 1 ∣}{2}))

Substitute back

u = \frac{4}{5} x

= \frac{1}{20} (- (\frac{ln \frac{4}{5} x + 1}{2} - \frac{ln \frac{4}{5} x - 1}{2}))

Simplify

\frac{1}{20} (- (\frac{ln \frac{4}{5} x + 1}{2} - \frac{ln \frac{4}{5} x - 1}{2})) : - \frac{1}{40} (ln \frac{4}{5} x + 1 - ln \frac{4}{5} x - 1)

= - \frac{1}{40} (ln \frac{4}{5} x + 1 - ln \frac{4}{5} x - 1)

Add a constant to the solution

= - \frac{1}{40} (ln \frac{4}{5} x + 1 - ln \frac{4}{5} x - 1) + C

\frac{d}{d x} (cos (2 x) - 2 sin (x))

\int_{2}^{3} \int_{2}^{6} (x y e^{x + y}) d y d x

f (x) = \frac{x - 3}{2}

2 x y y^{^{'}} = x^{2} - y^{2}

x \to 0 lim (csc (6 x))

What is the integral of 1/(16x^2-25) ?
The integral of 1/(16x^2-25) is -1/40 (ln| 4/5 x+1|-ln| 4/5 x-1|)+C