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

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

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

\int (\frac{16}{1 + x ^{2}}) d x

16 arctan (x) + C

Solution steps

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

Take the constant out:

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

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

Use the common integral:

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

= 16 arctan (x)

Add a constant to the solution

= 16 arctan (x) + C

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