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

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

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

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

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

Solution steps

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

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

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

Long division

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

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

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

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

\int 1 d x = x

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

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

Simplify

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

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

Add a constant to the solution

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

\int \frac{1 - 25 x ^{2}}{x ^{2}} d x

\int 3 \frac{cos ( x ) + sin ( x )}{sin ( 2 x )} d x

e^{2 x + 1}

f (x) = 5 x^{- 2}

\int (3 x - 2) e - x d x

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