What is the integral of 1/(16x^2-9) ?

The integral of 1/(16x^2-9) is -1/24 (ln| 4/3 x+1|-ln| 4/3 x-1|)+C

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integral of 1/(16x^2-9)

\int \frac{1}{16 x ^{2} - 9} d x

- \frac{1}{24} (ln \frac{4}{3} x + 1 - ln \frac{4}{3} x - 1) + C

Solution steps

\int \frac{1}{16 x ^{2} - 9} d x

Apply Integral Substitution

= \int \frac{1}{12 ( 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}{12} \cdot \int \frac{1}{u ^{2} - 1} d u

Factor

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

= \frac{1}{12} \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}{12} (- \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}{12} (- (\frac{ln ∣ u + 1 ∣}{2} - \frac{ln ∣ u - 1 ∣}{2}))

Substitute back

u = \frac{4}{3} x

= \frac{1}{12} (- (\frac{ln \frac{4}{3} x + 1}{2} - \frac{ln \frac{4}{3} x - 1}{2}))

Simplify

\frac{1}{12} (- (\frac{ln \frac{4}{3} x + 1}{2} - \frac{ln \frac{4}{3} x - 1}{2})) : - \frac{1}{24} (ln \frac{4}{3} x + 1 - ln \frac{4}{3} x - 1)

= - \frac{1}{24} (ln \frac{4}{3} x + 1 - ln \frac{4}{3} x - 1)

Add a constant to the solution

= - \frac{1}{24} (ln \frac{4}{3} x + 1 - ln \frac{4}{3} x - 1) + C

\frac{4 x + 3}{4 x - 3}

y^{^{''}} + 36 y = e^{- t}

\int (\frac{1}{x ^{9}}) d x

\int 4 x^{\frac{1}{2}} + 2 e^{x} + 3^{x} - 3 d x

\int \frac{x ^{2}}{( 2 + x ^{3} ) ^{2}} d x

What is the integral of 1/(16x^2-9) ?
The integral of 1/(16x^2-9) is -1/24 (ln| 4/3 x+1|-ln| 4/3 x-1|)+C