What is the integral of ((x-1)(x^2+4))/(x(x+1)) ?

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

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integral of ((x-1)(x^2+4))/(x(x+1))

\int \frac{( x - 1 ) ( x ^{2} + 4 )}{x ( x + 1 )} d x

\frac{x ^{2}}{2} - 2 x + 10 ln ∣ x + 1 ∣ - 4 ln ∣ x ∣ + C

Solution steps

\int \frac{( x - 1 ) ( x ^{2} + 4 )}{x ( x + 1 )} d x

Long division

\frac{( x - 1 ) ( x ^{2} + 4 )}{x ( x + 1 )} : x - 2 + \frac{6 x - 4}{x ^{2} + x}

= \int x - 2 + \frac{6 x - 4}{x ^{2} + x} 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 x d x - \int 2 d x + \int \frac{6 x - 4}{x ^{2} + x} d x

\int x d x = \frac{x ^{2}}{2}

\int 2 d x = 2 x

\int \frac{6 x - 4}{x ^{2} + x} d x = 10 ln ∣ x + 1 ∣ - 4 ln ∣ x ∣

= \frac{x ^{2}}{2} - 2 x + 10 ln ∣ x + 1 ∣ - 4 ln ∣ x ∣

Add a constant to the solution

= \frac{x ^{2}}{2} - 2 x + 10 ln ∣ x + 1 ∣ - 4 ln ∣ x ∣ + C

\int \frac{- 5 x ^{2} + 11 x - 10}{x ^{3} - x ^{2} - 6 x} d x

\frac{40 x ^{5}}{e ^{x}}

\frac{d}{d x} (\frac{2}{x - 7})

(4 x y + 3 y^{2} - x) d x + (x^{2} + 2 x y) d y = 0

y = \frac{x ^{2}}{x + 4}, (3, \frac{9}{7})

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