What is the integral of (2x-5)/(x^2+9) ?

The integral of (2x-5)/(x^2+9) is ln|(x^2)/9+1|-5/3 arctan(x/3)+C

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integral of (2x-5)/(x^2+9)

\int \frac{2 x - 5}{x ^{2} + 9} d x

ln \frac{x ^{2}}{9} + 1 - \frac{5}{3} arctan (\frac{x}{3}) + C

Solution steps

\int \frac{2 x - 5}{x ^{2} + 9} d x

Apply Integral Substitution

= \int \frac{6 u - 5}{3 ( u ^{2} + 1 )} d u

Take the constant out:

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

= \frac{1}{3} \cdot \int \frac{6 u - 5}{u ^{2} + 1} d u

Expand

\frac{6 u - 5}{u ^{2} + 1} : \frac{6 u}{u ^{2} + 1} - \frac{5}{u ^{2} + 1}

Apply the Sum Rule:

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

= \frac{1}{3} (\int \frac{6 u}{u ^{2} + 1} d u - \int \frac{5}{u ^{2} + 1} d u)

\int \frac{6 u}{u ^{2} + 1} d u = 3 ln u^{2} + 1

\int \frac{5}{u ^{2} + 1} d u = 5 arctan (u)

= \frac{1}{3} (3 ln u^{2} + 1 - 5 arctan (u))

Substitute back

u = \frac{x}{3}

= \frac{1}{3} (3 ln (\frac{x}{3})^{2} + 1 - 5 arctan (\frac{x}{3}))

Simplify

\frac{1}{3} (3 ln (\frac{x}{3})^{2} + 1 - 5 arctan (\frac{x}{3})) : ln \frac{x ^{2}}{9} + 1 - \frac{5}{3} arctan (\frac{x}{3})

= ln \frac{x ^{2}}{9} + 1 - \frac{5}{3} arctan (\frac{x}{3})

Add a constant to the solution

= ln \frac{x ^{2}}{9} + 1 - \frac{5}{3} arctan (\frac{x}{3}) + C

\frac{\partial}{\partial x} (6 x^{8} y^{6} + 2 x^{7} y^{5})

t^{3} y^{^{'}} + 4 t^{2} y = e^{- t}, y (1) = 0

x^{- 2}, 9

\int_{0}^{2} ∣ 2 x - 1 ∣ d x

e^{x^{2} - x}

What is the integral of (2x-5)/(x^2+9) ?
The integral of (2x-5)/(x^2+9) is ln|(x^2)/9+1|-5/3 arctan(x/3)+C