What is the integral of (17)/(x^2+1) ?

The integral of (17)/(x^2+1) is 17arctan(x)+C

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integral of (17)/(x^2+1)

\int \frac{17}{x ^{2} + 1} d x

17 arctan (x) + C

Solution steps

\int \frac{17}{x ^{2} + 1} d x

Take the constant out:

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

= 17 \cdot \int \frac{1}{x ^{2} + 1} d x

Use the common integral:

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

= 17 arctan (x)

Add a constant to the solution

= 17 arctan (x) + C

d er i v a t i v e 5 x (x - 1)

\int 2.4

x \to 0 lim (\frac{( ln ( x ) )}{x})

y^{^{''}} - 4 y^{^{'}} + 10 y = 0

\frac{\partial}{\partial x} (x^{\frac{4}{7}} y^{\frac{1}{7}} z^{\frac{2}{7}})

What is the integral of (17)/(x^2+1) ?
The integral of (17)/(x^2+1) is 17arctan(x)+C