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

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

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

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

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

Solution steps

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

Take the partial fraction of

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

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

\int \frac{1}{x} d x = ln ∣ x ∣

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

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

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

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

Simplify

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

Add a constant to the solution

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

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