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

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

\int x y d x

s l o p e (7, - 7), (5, - 3)

h \to 0 lim (\frac{( cos ( x + h ) - cos ( x ) )}{h})

n = 0 \sum \infty 3 x^{2 n + 1}

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