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))

integral

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

\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{1}{( cos ^{2} ( x ) ( 1 - tan ( x ) ) ^{2} )}

\frac{7 x + 2}{1 - 25 x ^{2}}

a^{2} (e^{- a})^{2} x^{4}

\frac{ln ( 4 x )}{x ^{4}}

\frac{1}{( 5 + x ^{2} )}

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