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

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

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

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

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

Solution steps

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

Apply Integral Substitution

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

Substitute back

u = \frac{x}{a}

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

Add a constant to the solution

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

\frac{1}{s ^{2} - 2 s + 4}

x \frac{d y}{d x} - 4 y = 0

e^{4 x}

\int \frac{1}{x ( 1 + 4 ln ^{2} ( x ) )} d x

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

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