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

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

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

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

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

Solution steps

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

Take the partial fraction of

\frac{3 x - 1}{x ^{2} ( x + 1 )} : \frac{4}{x} - \frac{1}{x ^{2}} - \frac{4}{x + 1}

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

\int \frac{4}{x} d x = 4 ln ∣ x ∣

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

\int \frac{4}{x + 1} d x = 4 ln ∣ x + 1 ∣

= 4 ln ∣ x ∣ - (- \frac{1}{x}) - 4 ln ∣ x + 1 ∣

Simplify

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

Add a constant to the solution

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

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

x \to 0 - lim (x [x] cot (x))

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

\int_{0}^{1} \frac{1}{x ^{- 2}} d x

\frac{d}{d x} (x^{2} cos (\frac{1}{x}))

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