What is the integral of (8x^3+8x^2+8)/(x^2+x) ?

The integral of (8x^3+8x^2+8)/(x^2+x) is 4x^2+8ln|x|-8ln|x+1|+C

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integral of (8x^3+8x^2+8)/(x^2+x)

\int \frac{8 x ^{3} + 8 x ^{2} + 8}{x ^{2} + x} d x

4 x^{2} + 8 ln ∣ x ∣ - 8 ln ∣ x + 1 ∣ + C

Solution steps

\int \frac{8 x ^{3} + 8 x ^{2} + 8}{x ^{2} + x} d x

Take the partial fraction of

\frac{8 x ^{3} + 8 x ^{2} + 8}{x ^{2} + x} : 8 x + \frac{8}{x} - \frac{8}{x + 1}

= \int 8 x + \frac{8}{x} - \frac{8}{x + 1} d 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 8 x d x + \int \frac{8}{x} d x - \int \frac{8}{x + 1} d x

\int 8 x d x = 4 x^{2}

\int \frac{8}{x} d x = 8 ln ∣ x ∣

\int \frac{8}{x + 1} d x = 8 ln ∣ x + 1 ∣

= 4 x^{2} + 8 ln ∣ x ∣ - 8 ln ∣ x + 1 ∣

Add a constant to the solution

= 4 x^{2} + 8 ln ∣ x ∣ - 8 ln ∣ x + 1 ∣ + C

\int \frac{x ln ( x )}{x} d x

f (x) = \frac{1}{x ^{5}}, (2, \frac{1}{32})

f (t) = 2 t^{2} - t

y^{^{'}} + \frac{1}{x + 1} y = x^{- 2}

f (x) = (3 x - 2) (x + 4), (1, 5)

What is the integral of (8x^3+8x^2+8)/(x^2+x) ?
The integral of (8x^3+8x^2+8)/(x^2+x) is 4x^2+8ln|x|-8ln|x+1|+C