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

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

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integral of (1-x^4)/(1-x)

integral

\frac{1 - x ^{4}}{1 - x}

\frac{x ^{4}}{4} + \frac{x ^{3}}{3} + \frac{x ^{2}}{2} + x + C

Solution steps

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

Expand

\frac{1 - x ^{4}}{1 - x} : \frac{1}{1 - x} - \frac{x ^{4}}{1 - 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}{1 - x} d x - \int \frac{x ^{4}}{1 - x} d x

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

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

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

Simplify

= \frac{x ^{4}}{4} + \frac{x ^{3}}{3} + \frac{x ^{2}}{2} + x

Add a constant to the solution

= \frac{x ^{4}}{4} + \frac{x ^{3}}{3} + \frac{x ^{2}}{2} + x + C

\frac{( cos ^{3} ( x ) )}{sin ^{4} ( x )}

(t^{2})^{2}

\frac{1}{( 2 x - x ^{2} )}

\frac{3 x ^{2} + 2 x}{x ^{6}}

1 + 2 x

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