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

The integral of (x-1)/(x^2+9) is 1/2 ln|x^2+9|-1/3 arctan(x/3)+C

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integral of (x-1)/(x^2+9)

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

\frac{1}{2} ln x^{2} + 9 - \frac{1}{3} arctan (\frac{x}{3}) + C

Solution steps

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

Expand

\frac{x - 1}{x ^{2} + 9} : \frac{x}{x ^{2} + 9} - \frac{1}{x ^{2} + 9}

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

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

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

\int \frac{1}{x ^{2} + 9} d x = \frac{1}{3} arctan (\frac{x}{3})

= \frac{1}{2} ln x^{2} + 9 - \frac{1}{3} arctan (\frac{x}{3})

Add a constant to the solution

= \frac{1}{2} ln x^{2} + 9 - \frac{1}{3} arctan (\frac{x}{3}) + C

y^{^{''}} + 4 y^{^{'}} + 5 y = 8

- 5 x^{2} + 4

\frac{d}{d x} (6 - 5 cos (\frac{π x}{15}))

\int \frac{x ^{4} + 4}{x ^{4}} d x

\frac{d}{d x} (x - 1 + x + 1)

What is the integral of (x-1)/(x^2+9) ?
The integral of (x-1)/(x^2+9) is 1/2 ln|x^2+9|-1/3 arctan(x/3)+C