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

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

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

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

\frac{x ^{4}}{4} + \frac{x ^{3}}{3} + \frac{x ^{2}}{2} + x + ln ∣ x - 1 ∣ + C

Solution steps

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

Long division

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

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

\int x^{3} d x = \frac{x ^{4}}{4}

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

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

\int 1 d x = x

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

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

Add a constant to the solution

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

\frac{6}{s ( s + 2 ) ^{2} ( s + 3 )}

\int \frac{1}{7 x ^{4}} d x

\int e^{x} \cdot sin (x - π) d x

x^{4} + 4 x^{2} - x

x \to - 1 lim (f (x) g (x))

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