What is the integral of 1/(4+t^2) ?

The integral of 1/(4+t^2) is 1/2 arctan(t/2)+C

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integral of 1/(4+t^2)

\int \frac{1}{4 + t ^{2}} d t

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

Solution steps

\int \frac{1}{4 + t ^{2}} d t

Apply Integral Substitution

= \int \frac{1}{2 ( 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}{2} \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}{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

\frac{\partial}{\partial x} (\frac{x - y}{x ^{2} + y ^{2} + 1})

f (x) = \frac{6 x + 14}{x + 7}

\int \frac{5}{sec ( x ) - 1} d x

\int \frac{1}{4 x - x ^{2}} d x

x^{2} - 4 \cdot x, x \cdot 0, [- 2, 4]

What is the integral of 1/(4+t^2) ?
The integral of 1/(4+t^2) is 1/2 arctan(t/2)+C