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

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

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integral of 1/(u^4+u)

\int \frac{1}{u ^{4} + u} d u

ln ∣ u ∣ - \frac{1}{3} ln ∣ u + 1 ∣ - \frac{1}{3} ln u^{2} - u + 1 + C

Solution steps

\int \frac{1}{u ^{4} + u} d u

Take the partial fraction of

\frac{1}{u ^{4} + u} : \frac{1}{u} - \frac{1}{3 ( u + 1 )} + \frac{- 2 u + 1}{3 ( u ^{2} - u + 1 )}

= \int \frac{1}{u} - \frac{1}{3 ( u + 1 )} + \frac{- 2 u + 1}{3 ( u ^{2} - u + 1 )} d u

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}{u} d u - \int \frac{1}{3 ( u + 1 )} d u + \int \frac{- 2 u + 1}{3 ( u ^{2} - u + 1 )} d u

\int \frac{1}{u} d u = ln ∣ u ∣

\int \frac{1}{3 ( u + 1 )} d u = \frac{1}{3} ln ∣ u + 1 ∣

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

= ln ∣ u ∣ - \frac{1}{3} ln ∣ u + 1 ∣ - \frac{1}{3} ln u^{2} - u + 1

Add a constant to the solution

= ln ∣ u ∣ - \frac{1}{3} ln ∣ u + 1 ∣ - \frac{1}{3} ln u^{2} - u + 1 + C

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

\int (x^{2} - 4 x) e^{x} d x

\int (5 x^{4} - 3) d x

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

\int [m^{2} ln (m)] d m

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