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

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

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integral of 8/(x^2+1)

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

8 arctan (x) + C

Solution steps

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

Use the common integral:

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

= 8 arctan (x)

Add a constant to the solution

= 8 arctan (x) + C

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

\frac{\partial}{\partial x} (x + 2 y - 1)

- x^{2} - 3 x, [- 6, - 1]

\frac{\partial}{\partial y} (2 y)

x y^{^{'}} + 4 y = 5 x