What is the integral of (12)/(1+t^2) ?

The integral of (12)/(1+t^2) is 12arctan(t)+C

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integral of (12)/(1+t^2)

\int \frac{12}{1 + t ^{2}} d t

12 arctan (t) + C

Solution steps

\int \frac{12}{1 + t ^{2}} d t

Take the constant out:

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

= 12 \cdot \int \frac{1}{1 + t ^{2}} d t

Use the common integral:

\int \frac{1}{1 + t ^{2}} d t = arctan (t)

= 12 arctan (t)

Add a constant to the solution

= 12 arctan (t) + C

- 5 sin (4 t)

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

\frac{s ^{3} + 5 s ^{2} + 9 s + 7}{s ^{2} + 3 s + 2}

\frac{\partial}{\partial x} (a x^{2})

x \to - 3 lim (\frac{x ^{2} - 4}{3 - x})

What is the integral of (12)/(1+t^2) ?
The integral of (12)/(1+t^2) is 12arctan(t)+C