What is the integral of (4x^2-5x-15)/(x^3-4x^2-5x) ?

The integral of (4x^2-5x-15)/(x^3-4x^2-5x) is -ln|x+1|+2ln|x-5|+3ln|x|+C

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integral of (4x^2-5x-15)/(x^3-4x^2-5x)

\int \frac{4 x ^{2} - 5 x - 15}{x ^{3} - 4 x ^{2} - 5 x} d x

- ln ∣ x + 1 ∣ + 2 ln ∣ x - 5 ∣ + 3 ln ∣ x ∣ + C

Solution steps

\int \frac{4 x ^{2} - 5 x - 15}{x ^{3} - 4 x ^{2} - 5 x} d x

Complete the square

4 x^{2} - 5 x - 15 : 4 (x - \frac{5}{8})^{2} - \frac{265}{16}

= \int \frac{4 ( x - \frac{5}{8} ) ^{2} - \frac{265}{16}}{x ^{3} - 4 x ^{2} - 5 x} d x

Apply u-substitution

: \int \frac{64 u ^{2} - 265}{16 ( ( u + \frac{5}{8} ) ^{3} - 4 ( u + \frac{5}{8} ) ^{2} - 5 ( u + \frac{5}{8} ) )} d u

= \int \frac{64 u ^{2} - 265}{16 ( ( u + \frac{5}{8} ) ^{3} - 4 ( u + \frac{5}{8} ) ^{2} - 5 ( u + \frac{5}{8} ) )} d u

Take the constant out:

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

= \frac{1}{16} \cdot \int \frac{64 u ^{2} - 265}{( u + \frac{5}{8} ) ^{3} - 4 ( u + \frac{5}{8} ) ^{2} - 5 ( u + \frac{5}{8} )} d u

Apply u-substitution

: \int \frac{64 v ^{2} - 80 v - 240}{v ^{3} - 4 v ^{2} - 5 v} d v

= \frac{1}{16} \cdot \int \frac{64 v ^{2} - 80 v - 240}{v ^{3} - 4 v ^{2} - 5 v} d v

Expand

\frac{64 v ^{2} - 80 v - 240}{v ^{3} - 4 v ^{2} - 5 v} : \frac{64 v ^{2}}{v ^{3} - 4 v ^{2} - 5 v} - \frac{80 v}{v ^{3} - 4 v ^{2} - 5 v} - \frac{240}{v ^{3} - 4 v ^{2} - 5 v}

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}{16} (\int \frac{64 v ^{2}}{v ^{3} - 4 v ^{2} - 5 v} d v - \int \frac{80 v}{v ^{3} - 4 v ^{2} - 5 v} d v - \int \frac{240}{v ^{3} - 4 v ^{2} - 5 v} d v)

\int \frac{64 v ^{2}}{v ^{3} - 4 v ^{2} - 5 v} d v = 64 (\frac{1}{6} ln ∣ v + 1 ∣ + \frac{5}{6} ln ∣ v - 5 ∣)

\int \frac{80 v}{v ^{3} - 4 v ^{2} - 5 v} d v = 80 (- \frac{1}{6} ln ∣ v + 1 ∣ + \frac{1}{6} ln ∣ v - 5 ∣)

\int \frac{240}{v ^{3} - 4 v ^{2} - 5 v} d v = 240 (- \frac{1}{5} ln ∣ v ∣ + \frac{1}{6} ln ∣ v + 1 ∣ + \frac{1}{30} ln ∣ v - 5 ∣)

= \frac{1}{16} (64 (\frac{1}{6} ln ∣ v + 1 ∣ + \frac{5}{6} ln ∣ v - 5 ∣) - 80 (- \frac{1}{6} ln ∣ v + 1 ∣ + \frac{1}{6} ln ∣ v - 5 ∣) - 240 (- \frac{1}{5} ln ∣ v ∣ + \frac{1}{6} ln ∣ v + 1 ∣ + \frac{1}{30} ln ∣ v - 5 ∣))

Substitute back

= \frac{1}{16} (64 (\frac{1}{6} ln x - \frac{5}{8} + \frac{5}{8} + 1 + \frac{5}{6} ln x - \frac{5}{8} + \frac{5}{8} - 5) - 80 (- \frac{1}{6} ln x - \frac{5}{8} + \frac{5}{8} + 1 + \frac{1}{6} ln x - \frac{5}{8} + \frac{5}{8} - 5) - 240 (- \frac{1}{5} ln x - \frac{5}{8} + \frac{5}{8} + \frac{1}{6} ln x - \frac{5}{8} + \frac{5}{8} + 1 + \frac{1}{30} ln x - \frac{5}{8} + \frac{5}{8} - 5))

Simplify

\frac{1}{16} (64 (\frac{1}{6} ln x - \frac{5}{8} + \frac{5}{8} + 1 + \frac{5}{6} ln x - \frac{5}{8} + \frac{5}{8} - 5) - 80 (- \frac{1}{6} ln x - \frac{5}{8} + \frac{5}{8} + 1 + \frac{1}{6} ln x - \frac{5}{8} + \frac{5}{8} - 5) - 240 (- \frac{1}{5} ln x - \frac{5}{8} + \frac{5}{8} + \frac{1}{6} ln x - \frac{5}{8} + \frac{5}{8} + 1 + \frac{1}{30} ln x - \frac{5}{8} + \frac{5}{8} - 5)) : - ln ∣ x + 1 ∣ + 2 ln ∣ x - 5 ∣ + 3 ln ∣ x ∣

= - ln ∣ x + 1 ∣ + 2 ln ∣ x - 5 ∣ + 3 ln ∣ x ∣

Add a constant to the solution

= - ln ∣ x + 1 ∣ + 2 ln ∣ x - 5 ∣ + 3 ln ∣ x ∣ + C

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What is the integral of (4x^2-5x-15)/(x^3-4x^2-5x) ?
The integral of (4x^2-5x-15)/(x^3-4x^2-5x) is -ln|x+1|+2ln|x-5|+3ln|x|+C