What is the integral of (5x^2-x-3)/(x^2(x+1)) ?

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

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integral of (5x^2-x-3)/(x^2(x+1))

\int \frac{5 x ^{2} - x - 3}{x ^{2} ( x + 1 )} d x

2 ln ∣ x ∣ + \frac{3}{x} + 3 ln ∣ x + 1 ∣ + C

Solution steps

\int \frac{5 x ^{2} - x - 3}{x ^{2} ( x + 1 )} d x

Take the partial fraction of

\frac{5 x ^{2} - x - 3}{x ^{2} ( x + 1 )} : \frac{2}{x} - \frac{3}{x ^{2}} + \frac{3}{x + 1}

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

\int \frac{2}{x} d x = 2 ln ∣ x ∣

\int \frac{3}{x ^{2}} d x = - \frac{3}{x}

\int \frac{3}{x + 1} d x = 3 ln ∣ x + 1 ∣

= 2 ln ∣ x ∣ - (- \frac{3}{x}) + 3 ln ∣ x + 1 ∣

Simplify

= 2 ln ∣ x ∣ + \frac{3}{x} + 3 ln ∣ x + 1 ∣

Add a constant to the solution

= 2 ln ∣ x ∣ + \frac{3}{x} + 3 ln ∣ x + 1 ∣ + C

x \to 0 + lim (x^{2} (ln (x)))

\frac{d}{d x} (sec (x - \frac{π}{4}))

\frac{\partial}{\partial x} (7 e^{x y + 5})

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

(- \frac{10}{63}, \frac{1}{3}), (\frac{2}{3}, - \frac{11}{45})

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