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

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

(tan (x)) (x)

\frac{\partial}{\partial x} (ln (2 x + 5 y))

y^{^{''}} + 2 y^{^{'}} + \frac{3}{4} y = 0

\int 3 - \frac{1}{x ^{4}} d x

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