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

The integral of 1/((x^2+7)) is 1/(sqrt(7))arctan(x/(sqrt(7)))+C

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integral of 1/((x^2+7))

\int \frac{1}{( x ^{2} + 7 )} d x

\frac{1}{7} arctan (\frac{x}{7}) + 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

Use the common integral:

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

= \frac{1}{7} arctan (u)

Substitute back

u = \frac{x}{7}

= \frac{1}{7} arctan (\frac{x}{7})

Add a constant to the solution

= \frac{1}{7} arctan (\frac{x}{7}) + C

y = 1 - 4 x, at x = - 1

x \to 9 - lim (\frac{x}{x - 9})

\int - ln (cos (x)) d x

x \to 0 lim (2 x - x ln (x))

\frac{d}{d x} (3 x^{2} + 2 x + 5)

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