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

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

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

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

- ln ∣ x ∣ + ln x^{2} + 1 + C

Solution steps

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

Take the partial fraction of

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

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

\int \frac{1}{x} d x = ln ∣ x ∣

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

= - ln ∣ x ∣ + ln x^{2} + 1

Add a constant to the solution

= - ln ∣ x ∣ + ln x^{2} + 1 + C

\int_{0}^{1} x e^{3 x^{2}} d x

tan (t) y^{^{'}} - y = 1, y (\frac{π}{6}) = 3

\frac{x}{x ^{2} - x + 1}

y^{^{''}} - 2 y^{^{'}} + y = e^{t}

\frac{d}{d x} (\frac{7 x}{x + 4})

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