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

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

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integral of 1/(2+t^2)

\int \frac{1}{2 + t ^{2}} d t

\frac{1}{2} arctan (\frac{t}{2}) + C

Solution steps

\int \frac{1}{2 + t ^{2}} d t

Apply Integral Substitution

= \int \frac{1}{2 ( 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}{2} \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}{2} arctan (u)

Substitute back

u = \frac{t}{2}

= \frac{1}{2} arctan (\frac{t}{2})

Add a constant to the solution

= \frac{1}{2} arctan (\frac{t}{2}) + C

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