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

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

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integral of (x^4+x^2-1)/(x-1)

\int \frac{x ^{4} + x ^{2} - 1}{x - 1} d x

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

Solution steps

\int \frac{x ^{4} + x ^{2} - 1}{x - 1} d x

Expand

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

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{x ^{4}}{x - 1} d x + \int \frac{x ^{2}}{x - 1} d x - \int \frac{1}{x - 1} d x

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

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

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

Simplify

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

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

Add a constant to the solution

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

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

x \to a lim (\frac{ln ( x ) - ln ( a )}{x - a})

\int (\frac{sec ( x )}{tan ( x )})^{4} d x

\frac{d y}{d x} = e^{4 x - 3 y}, y (0) = 2

\int \frac{sin ^{5} ( x )}{cos ( x )} d x

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