What is the integral of (5x+3)/(x^2-4) ?

The integral of (5x+3)/(x^2-4) is 1/2 (5ln|(x^2)/4-1|-3(1/2 ln| x/2+1|-1/2 ln| x/2-1|))+C

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integral of (5x+3)/(x^2-4)

\int \frac{5 x + 3}{x ^{2} - 4} d x

\frac{1}{2} (5 ln \frac{x ^{2}}{4} - 1 - 3 (\frac{1}{2} ln \frac{x}{2} + 1 - \frac{1}{2} ln \frac{x}{2} - 1)) + C

Solution steps

\int \frac{5 x + 3}{x ^{2} - 4} d x

Apply Integral Substitution

= \int \frac{10 u + 3}{2 ( 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}{2} \cdot \int \frac{10 u + 3}{u ^{2} - 1} d u

Expand

\frac{10 u + 3}{u ^{2} - 1} : \frac{10 u}{u ^{2} - 1} + \frac{3}{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}{2} (\int \frac{10 u}{u ^{2} - 1} d u + \int \frac{3}{u ^{2} - 1} d u)

\int \frac{10 u}{u ^{2} - 1} d u = 5 ln u^{2} - 1

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

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

Substitute back

u = \frac{x}{2}

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

(\frac{x}{2})^{2} = \frac{x ^{2}}{4}

= \frac{1}{2} (5 ln \frac{x ^{2}}{4} - 1 - 3 (\frac{1}{2} ln \frac{x}{2} + 1 - \frac{1}{2} ln \frac{x}{2} - 1))

Add a constant to the solution

= \frac{1}{2} (5 ln \frac{x ^{2}}{4} - 1 - 3 (\frac{1}{2} ln \frac{x}{2} + 1 - \frac{1}{2} ln \frac{x}{2} - 1)) + C

What is the integral of (5x+3)/(x^2-4) ?
The integral of (5x+3)/(x^2-4) is 1/2 (5ln|(x^2)/4-1|-3(1/2 ln| x/2+1|-1/2 ln| x/2-1|))+C