What is the integral of 9/(x^2+1) ?

The integral of 9/(x^2+1) is 9arctan(x)+C

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integral of 9/(x^2+1)

\int \frac{9}{x ^{2} + 1} d x

9 arctan (x) + C

Solution steps

\int \frac{9}{x ^{2} + 1} d x

Take the constant out:

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

= 9 \cdot \int \frac{1}{x ^{2} + 1} d x

Use the common integral:

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

= 9 arctan (x)

Add a constant to the solution

= 9 arctan (x) + C

(\frac{x}{1 - x})^{^{'}}

y^{^{''}} - 8 y^{^{'}} + 20 y = 200 x^{2} - 52 x e^{x}

\frac{\partial}{\partial x} (7 x + 6 y + 7 x y)

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x^{2} - 4 x + 1, 3 x - 5, 1, 4