What is the integral of ((5x+3))/((x^2-9)) ?

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

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integral of ((5x+3))/((x^2-9))

\int \frac{( 5 x + 3 )}{( x ^{2} - 9 )} d x

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

Solution steps

\int \frac{5 x + 3}{x ^{2} - 9} d x

Apply Integral Substitution

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

Expand

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

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

\int \frac{5 u}{u ^{2} - 1} d u = \frac{5}{2} ln u^{2} - 1

\int \frac{1}{u ^{2} - 1} d u = - \frac{1}{2} ln ∣ u + 1 ∣ + \frac{1}{2} ln ∣ u - 1 ∣

= \frac{5}{2} ln u^{2} - 1 - \frac{1}{2} ln ∣ u + 1 ∣ + \frac{1}{2} ln ∣ u - 1 ∣

Substitute back

u = \frac{x}{3}

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

(\frac{x}{3})^{2} = \frac{x ^{2}}{9}

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

Add a constant to the solution

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

\frac{d y}{d x} = \frac{( - y ( x + y + 1 ) )}{x + 2 y}

\int_{- 1}^{1} (1 + 1 - x^{2}) d x

t^{2} y^{^{''}} + 19 t y^{^{'}} + 162 y = 0

3 cos (2 u)

n = 2 \sum \infty \frac{1}{n ^{2} + n}

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