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

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

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

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

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

Solution steps

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

Long division

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

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

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

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

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

Simplify

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

Add a constant to the solution

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

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