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

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

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integral of 1/(7+x^2)

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

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

Solution steps

\int \frac{1}{7 + 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 = 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

(e^{x} + 1) \frac{d y}{d x} = y - y e^{x}

x \to 0 lim (\frac{cot ( x )}{x})

y = 3 - 2 x^{2}, (- 4, - 29)

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

x \to - \infty lim ((x)^{2})

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