What is the integral of (4*x+3)/(x^2-25) ?

The integral of (4*x+3)/(x^2-25) is 1/5 (10ln|(x^2)/(25)-1|-3(1/2 ln| x/5+1|-1/2 ln| x/5-1|))+C

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integral of (4*x+3)/(x^2-25)

\int \frac{4 \cdot x + 3}{x ^{2} - 25} d x

\frac{1}{5} (10 ln \frac{x ^{2}}{25} - 1 - 3 (\frac{1}{2} ln \frac{x}{5} + 1 - \frac{1}{2} ln \frac{x}{5} - 1)) + C

Solution steps

\int \frac{4 x + 3}{x ^{2} - 25} d x

Apply Integral Substitution

= \int \frac{20 u + 3}{5 ( 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}{5} \cdot \int \frac{20 u + 3}{u ^{2} - 1} d u

Expand

\frac{20 u + 3}{u ^{2} - 1} : \frac{20 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}{5} (\int \frac{20 u}{u ^{2} - 1} d u + \int \frac{3}{u ^{2} - 1} d u)

\int \frac{20 u}{u ^{2} - 1} d u = 10 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}{5} (10 ln u^{2} - 1 - 3 (\frac{1}{2} ln ∣ u + 1 ∣ - \frac{1}{2} ln ∣ u - 1 ∣))

Substitute back

u = \frac{x}{5}

= \frac{1}{5} (10 ln (\frac{x}{5})^{2} - 1 - 3 (\frac{1}{2} ln \frac{x}{5} + 1 - \frac{1}{2} ln \frac{x}{5} - 1))

(\frac{x}{5})^{2} = \frac{x ^{2}}{25}

= \frac{1}{5} (10 ln \frac{x ^{2}}{25} - 1 - 3 (\frac{1}{2} ln \frac{x}{5} + 1 - \frac{1}{2} ln \frac{x}{5} - 1))

Add a constant to the solution

= \frac{1}{5} (10 ln \frac{x ^{2}}{25} - 1 - 3 (\frac{1}{2} ln \frac{x}{5} + 1 - \frac{1}{2} ln \frac{x}{5} - 1)) + C

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What is the integral of (4*x+3)/(x^2-25) ?
The integral of (4*x+3)/(x^2-25) is 1/5 (10ln|(x^2)/(25)-1|-3(1/2 ln| x/5+1|-1/2 ln| x/5-1|))+C