What is the integral of (16)/(1+t^2) ?

The integral of (16)/(1+t^2) is 16arctan(t)+C

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integral of (16)/(1+t^2)

\int \frac{16}{1 + t ^{2}} d t

16 arctan (t) + C

Solution steps

\int \frac{16}{1 + t ^{2}} d t

Take the constant out:

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

= 16 \cdot \int \frac{1}{1 + t ^{2}} d t

Use the common integral:

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

= 16 arctan (t)

Add a constant to the solution

= 16 arctan (t) + C

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What is the integral of (16)/(1+t^2) ?
The integral of (16)/(1+t^2) is 16arctan(t)+C