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))

integral

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

\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

(1 - x)^{2}

\frac{x ^{2}}{( x + 2 ) ^{3}}

x^{2} e^{- x^{2}}

\frac{sin ^{4} ( x )}{cos ^{2} ( x )}

\frac{cos ( x )}{x} + \frac{sin ( x )}{x ^{2}}

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