What is the integral of 8/(1+x^2) ?

The integral of 8/(1+x^2) is 8arctan(x)+C

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integral of 8/(1+x^2)

\int \frac{8}{1 + x ^{2}} d x

8 arctan (x) + C

Solution steps

\int \frac{8}{1 + x ^{2}} d x

Take the constant out:

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

= 8 \cdot \int \frac{1}{1 + x ^{2}} d x

Use the common integral:

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

= 8 arctan (x)

Add a constant to the solution

= 8 arctan (x) + C

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