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

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

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

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

16 arctan (x) + C

Solution steps

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

Use the common integral:

\int \frac{1}{x ^{2} + 1} 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)/(x^2+1) ?
The integral of (16)/(x^2+1) is 16arctan(x)+C