What is the integral of (8x^2-5x+1)/(x^2+x) ?

The integral of (8x^2-5x+1)/(x^2+x) is (8x^2)/(x+1)-14ln|x+1|-8/(x+1)+ln|x|+C

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integral of (8x^2-5x+1)/(x^2+x)

\int \frac{8 x ^{2} - 5 x + 1}{x ^{2} + x} d x

\frac{8 x ^{2}}{x + 1} - 14 ln ∣ x + 1 ∣ - \frac{8}{x + 1} + ln ∣ x ∣ + C

Solution steps

\int \frac{8 x ^{2} - 5 x + 1}{x ^{2} + x} d x

Expand

\frac{8 x ^{2} - 5 x + 1}{x ^{2} + x} : \frac{8 x ^{2}}{x ^{2} + x} - \frac{5 x}{x ^{2} + x} + \frac{1}{x ^{2} + x}

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{8 x ^{2}}{x ^{2} + x} d x - \int \frac{5 x}{x ^{2} + x} d x + \int \frac{1}{x ^{2} + x} d x

\int \frac{8 x ^{2}}{x ^{2} + x} d x = 8 (\frac{x ^{2}}{x + 1} - ln ∣ x + 1 ∣ - \frac{1}{x + 1})

\int \frac{5 x}{x ^{2} + x} d x = 5 ln ∣ x + 1 ∣

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

= 8 (\frac{x ^{2}}{x + 1} - ln ∣ x + 1 ∣ - \frac{1}{x + 1}) - 5 ln ∣ x + 1 ∣ + ln ∣ x ∣ - ln ∣ x + 1 ∣

Simplify

8 (\frac{x ^{2}}{x + 1} - ln ∣ x + 1 ∣ - \frac{1}{x + 1}) - 5 ln ∣ x + 1 ∣ + ln ∣ x ∣ - ln ∣ x + 1 ∣ : \frac{8 x ^{2}}{x + 1} - 14 ln ∣ x + 1 ∣ - \frac{8}{x + 1} + ln ∣ x ∣

= \frac{8 x ^{2}}{x + 1} - 14 ln ∣ x + 1 ∣ - \frac{8}{x + 1} + ln ∣ x ∣

Add a constant to the solution

= \frac{8 x ^{2}}{x + 1} - 14 ln ∣ x + 1 ∣ - \frac{8}{x + 1} + ln ∣ x ∣ + C

\frac{d}{d x} (6 cos (x) + 6 x sin (x))

x \to 4 lim (4 + 2)

\int_{- 1}^{3} 1 d x

\frac{d}{d x} (- 9 (3 x^{4} - x^{3} + 2 x)^{5})

5 x^{2} + 2 x y - 3 y^{2} + 3 = 0, (2, 1)

What is the integral of (8x^2-5x+1)/(x^2+x) ?
The integral of (8x^2-5x+1)/(x^2+x) is (8x^2)/(x+1)-14ln|x+1|-8/(x+1)+ln|x|+C