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

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

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

\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)) + C

Solution steps

\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 )}

= \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

= \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))

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

Add a constant to the solution

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

\int_{0}^{1} π (x^{2} - x^{4}) d x

\frac{\partial}{\partial x} (y \cdot arcsin (\frac{x}{y}))

6 t^{2} - t

f (x) = x^{2} tan (x)

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

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