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

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

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integral of 1/(x^2+c)

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

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

Solution steps

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

x^{2} + c = c (\frac{x ^{2}}{c} + 1)

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

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

Apply Integral Substitution

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

Substitute back

u = \frac{1}{c} x

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

Simplify

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

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

Add a constant to the solution

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

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