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

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

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integral of 1/(49-x^2)

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

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

Solution steps

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

Apply Integral Substitution

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

Substitute back

u = \frac{x}{7}

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

Simplify

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

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

Add a constant to the solution

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

\frac{\partial}{\partial x} ((x^{2} + y) e^{3 x + 4 y})

\frac{d y}{d x} = 3 y + y^{4}

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

\frac{d}{d y} (\frac{x}{x ^{2} + y ^{2}})

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