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

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

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

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

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

Solution steps

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

Take the partial fraction of

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

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

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

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

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

Simplify

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

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

Add a constant to the solution

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

\int 7 x^{3} e^{5 x} d x

(1 + x)^{2}

x (\frac{d y}{d x}) - y = x^{2}

n = 0 \sum \infty \frac{2 ^{n}}{5 ^{n}}

x \to \infty - lim (\frac{1}{x})

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