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

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

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

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

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

Solution steps

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

Apply Integral Substitution

= \int \frac{1}{c ( 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}{c} \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}{c} arctan (u)

Substitute back

u = \frac{x}{c}

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

Add a constant to the solution

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

\frac{d x}{d t} + 5 t^{5} x^{7} + \frac{x}{t} = 0

x \to \infty lim (- 2 x^{5})

\int 2 x y d x

x \to 1 lim (1 + \frac{1}{x})

\int_{- 1}^{1} \frac{2 x}{( x + 6 ) ^{2}} d x

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