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

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

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

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

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

Solution steps

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

Expand

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

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

\int \frac{x ^{4}}{x ^{3} + x} d x = \frac{1}{2} (x^{2} + 1 - ln x^{2} + 1)

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

Simplify

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

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

Add a constant to the solution

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

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