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

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

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

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

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

Solution steps

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

Take the partial fraction of

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

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

\int \frac{1}{x} d x = ln ∣ x ∣

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

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

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

Add a constant to the solution

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

\int \frac{1}{x ^{2} + y ^{2}} d x

\frac{9}{s ^{2} + 9 s + 9} \cdot \frac{1}{s}

f (x) = x^{4} + 9 x^{2} - x, at x = 1

n = 3 \sum \infty (\frac{4}{7})^{n}

\int \frac{1}{cos ^{2} ( 4 x )} d x

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