What is the integral of 1/(1-49x^2) ?

The integral of 1/(1-49x^2) is 1/14 (ln|7x+1|-ln|7x-1|)+C

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integral of 1/(1-49x^2)

\int \frac{1}{1 - 49 x ^{2}} d x

\frac{1}{14} (ln ∣ 7 x + 1 ∣ - ln ∣ 7 x - 1 ∣) + C

Solution steps

\int \frac{1}{1 - 49 x ^{2}} d x

Apply Integral Substitution

= \int \frac{1}{7 ( 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}{7} \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}{7} (\frac{ln ∣ u + 1 ∣}{2} - \frac{ln ∣ u - 1 ∣}{2})

Substitute back

u = 7 x

= \frac{1}{7} (\frac{ln ∣ 7 x + 1 ∣}{2} - \frac{ln ∣ 7 x - 1 ∣}{2})

Simplify

\frac{1}{7} (\frac{ln ∣ 7 x + 1 ∣}{2} - \frac{ln ∣ 7 x - 1 ∣}{2}) : \frac{1}{14} (ln ∣ 7 x + 1 ∣ - ln ∣ 7 x - 1 ∣)

= \frac{1}{14} (ln ∣ 7 x + 1 ∣ - ln ∣ 7 x - 1 ∣)

Add a constant to the solution

= \frac{1}{14} (ln ∣ 7 x + 1 ∣ - ln ∣ 7 x - 1 ∣) + C

\int \frac{2}{x - 3} d x

\int \frac{x}{x + 4} d x

\frac{d}{d x} ((x^{2} - 3) (x^{3} - 2 x))

f (x) = \frac{2 x}{x ^{2} - 7}, (3, 3)

\frac{d}{d x} (- 11 e^{x} sin (2 x) - 2 e^{x} cos (2 x))

What is the integral of 1/(1-49x^2) ?
The integral of 1/(1-49x^2) is 1/14 (ln|7x+1|-ln|7x-1|)+C