What is the integral of (6x^2+34x-7)/(x^3+7x^2) ?

The integral of (6x^2+34x-7)/(x^3+7x^2) is 5ln|x|+ln|x+7|+1/x+C

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integral of (6x^2+34x-7)/(x^3+7x^2)

\int \frac{6 x ^{2} + 34 x - 7}{x ^{3} + 7 x ^{2}} d x

5 ln ∣ x ∣ + ln ∣ x + 7 ∣ + \frac{1}{x} + C

Solution steps

\int \frac{6 x ^{2} + 34 x - 7}{x ^{3} + 7 x ^{2}} d x

Expand

\frac{6 x ^{2} + 34 x - 7}{x ^{3} + 7 x ^{2}} : \frac{6 x ^{2}}{x ^{3} + 7 x ^{2}} + \frac{34 x}{x ^{3} + 7 x ^{2}} - \frac{7}{x ^{3} + 7 x ^{2}}

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{6 x ^{2}}{x ^{3} + 7 x ^{2}} d x + \int \frac{34 x}{x ^{3} + 7 x ^{2}} d x - \int \frac{7}{x ^{3} + 7 x ^{2}} d x

\int \frac{6 x ^{2}}{x ^{3} + 7 x ^{2}} d x = 6 ln ∣ x + 7 ∣

\int \frac{34 x}{x ^{3} + 7 x ^{2}} d x = 34 (\frac{1}{7} ln ∣ x ∣ - \frac{1}{7} ln ∣ x + 7 ∣)

\int \frac{7}{x ^{3} + 7 x ^{2}} d x = - \frac{1}{7} ln ∣ x ∣ - \frac{1}{x} + \frac{1}{7} ln ∣ x + 7 ∣

= 6 ln ∣ x + 7 ∣ + 34 (\frac{1}{7} ln ∣ x ∣ - \frac{1}{7} ln ∣ x + 7 ∣) - (- \frac{1}{7} ln ∣ x ∣ - \frac{1}{x} + \frac{1}{7} ln ∣ x + 7 ∣)

Simplify

6 ln ∣ x + 7 ∣ + 34 (\frac{1}{7} ln ∣ x ∣ - \frac{1}{7} ln ∣ x + 7 ∣) - (- \frac{1}{7} ln ∣ x ∣ - \frac{1}{x} + \frac{1}{7} ln ∣ x + 7 ∣) : 5 ln ∣ x ∣ + ln ∣ x + 7 ∣ + \frac{1}{x}

= 5 ln ∣ x ∣ + ln ∣ x + 7 ∣ + \frac{1}{x}

Add a constant to the solution

= 5 ln ∣ x ∣ + ln ∣ x + 7 ∣ + \frac{1}{x} + C

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What is the integral of (6x^2+34x-7)/(x^3+7x^2) ?
The integral of (6x^2+34x-7)/(x^3+7x^2) is 5ln|x|+ln|x+7|+1/x+C