What is the integral of 1/(x^2-7) ?

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

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integral of 1/(x^2-7)

\int \frac{1}{x ^{2} - 7} d x

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

Solution steps

\int \frac{1}{x ^{2} - 7} 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

Factor

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

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

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

Add a constant to the solution

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

- 4 y^{^{''}} + 16 y = 0, y (1) = 2, y^{^{'}} (2) = 1

y = x^{3} (x - 1)^{2}

f (x) = 6 sin^{3} (x)

\int 8 x^{2} + 6 + \frac{9}{x ^{2} + 1} d x

y^{^{''}} + 16 y^{^{'}} + 73 y = 0

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