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

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

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integral of 1/(c+x^2)

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

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

Solution steps

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

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

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

Apply Integral Substitution

= \frac{1}{c} \cdot \int \frac{c}{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}{c} c \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}{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/(c+x^2) ?
The integral of 1/(c+x^2) is 1/(sqrt(c))arctan(sqrt(1/c)x)+C