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

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

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

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

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

Solution steps

\int \frac{1}{25 x ^{2} - 16} 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{5}{4} x

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

Simplify

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

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

Add a constant to the solution

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

\int \frac{4 x}{2 x ^{2} + 1} d x

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\int 5 sin^{3} (2 x co) s^{3} 2 x d x

y^{^{''}} + 4 y = 4 sin (2 t)

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