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

The integral of (4x-1)/(x^3+2x^2+x) is -ln|x|+ln|x+1|-5/(x+1)+C

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integral of (4x-1)/(x^3+2x^2+x)

\int \frac{4 x - 1}{x ^{3} + 2 x ^{2} + x} d x

- ln ∣ x ∣ + ln ∣ x + 1 ∣ - \frac{5}{x + 1} + C

Solution steps

\int \frac{4 x - 1}{x ^{3} + 2 x ^{2} + x} d x

Take the partial fraction of

\frac{4 x - 1}{x ^{3} + 2 x ^{2} + x} : - \frac{1}{x} + \frac{1}{x + 1} + \frac{5}{( x + 1 ) ^{2}}

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

\int \frac{1}{x} d x = ln ∣ x ∣

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

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

= - ln ∣ x ∣ + ln ∣ x + 1 ∣ - \frac{5}{x + 1}

Add a constant to the solution

= - ln ∣ x ∣ + ln ∣ x + 1 ∣ - \frac{5}{x + 1} + C

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

\int sin (3 t) + cos (t) d t

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

x^{2} + y^{2} + 2 x y y^{^{'}} = 0

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

What is the integral of (4x-1)/(x^3+2x^2+x) ?
The integral of (4x-1)/(x^3+2x^2+x) is -ln|x|+ln|x+1|-5/(x+1)+C