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

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

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integral of 1/(x^2+a)

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

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

Solution steps

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

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

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

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{a} \cdot \int \frac{1}{\frac{x ^{2}}{a} + 1} d x

Apply Integral Substitution

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

Substitute back

u = \frac{1}{a} x

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

Simplify

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

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

Add a constant to the solution

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

\frac{d y}{d x} = 2 + 3 + y - 2 x

\frac{d}{d x} ((2 x + 5) (3 x - 4))

sin (2 x + 2)

\int co tan (\frac{x}{3}) d x

y^{^{''}} + 9 y = 2 cos (3 t)

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