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

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

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

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

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

Solution steps

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

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

Add a constant to the solution

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

f (x) = x^{2} - 2 x, (- 2, 8)

\frac{x ^{2} - 1}{x}

(3 x^{2} - x - 4) (- x^{2} + 2)

x \to 0 lim (6 x - ln (6 x + 1))

y^{^{''}} + 16 y = t e^{t} cos (4 t)

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