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

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

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

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

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

Solution steps

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

Take the partial fraction of

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

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

\int 1 d x = x

\int \frac{2}{x} d x = 2 ln ∣ x ∣

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

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

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

Simplify

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

Add a constant to the solution

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

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

\frac{d ^{2} y}{d x ^{2}} = 49 y

\frac{d}{d x} (6^{4} x^{3} + x^{5} - 180 4 x^{2})

\frac{d}{d x} (\frac{- e ^{- x} x - e ^{- x}}{x ^{2}})

\int_{- 7}^{7} (1 + 49 - x^{2}) d x

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