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

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

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

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

ln ∣ x - 1 ∣ - \frac{1}{2} ln x^{2} + 2 x + 2 - arctan (x + 1) + C

Solution steps

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

Take the partial fraction of

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

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

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

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

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

Add a constant to the solution

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

\frac{d}{d y} (x^{y^{z}})

f (x) = \frac{5}{1 - 9 x}

f (x) = 6 e^{x} + 4 x^{2}

\frac{\partial}{\partial y} (cos (12 x) sin (3 y))

\int \frac{1}{9 - y ^{2}} d y

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