What is the integral of 1/(1-4x^2) ?

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

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integral of 1/(1-4x^2)

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

\frac{1}{4} (ln ∣ 2 x + 1 ∣ - ln ∣ 2 x - 1 ∣) + C

Solution steps

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

Apply Integral Substitution

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

Take the constant out:

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

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

Use the common integral:

\int \frac{1}{1 - u ^{2}} d u = \frac{ln ∣ u + 1 ∣}{2} - \frac{ln ∣ u - 1 ∣}{2}

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

Substitute back

u = 2 x

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

Simplify

\frac{1}{2} (\frac{ln ∣ 2 x + 1 ∣}{2} - \frac{ln ∣ 2 x - 1 ∣}{2}) : \frac{1}{4} (ln ∣ 2 x + 1 ∣ - ln ∣ 2 x - 1 ∣)

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

Add a constant to the solution

= \frac{1}{4} (ln ∣ 2 x + 1 ∣ - ln ∣ 2 x - 1 ∣) + C

f (x) = - 5 - 3 x^{2}, - 53, at x = 4

(- 3.2) (7.6)

\int x^{2} (x^{3} - 1)^{10} d x

x \to - 3 lim (\frac{7}{x ^{7}})

\int x 2 d x

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