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

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

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integral of 9/(1+x^2)

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

9 arctan (x) + C

Solution steps

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

Use the common integral:

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

= 9 arctan (x)

Add a constant to the solution

= 9 arctan (x) + C

\int (\frac{5}{x} - \frac{3}{x ^{4}}) d x

y = \frac{2 x}{9 - tan ( x )}

\frac{d}{d x} (- π^{2} sin (π x))

\int_{y}^{1} 3 x d x

f (4 k) = x^{2} - 2 x