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

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

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

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

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

Solution steps

\int \frac{4 x}{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

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

Take the partial fraction of

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

= 4 \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

= 4 (- \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))

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

Add a constant to the solution

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

\int_{0}^{t} t^{2} e^{- 5 t} d t

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\frac{d}{d t} (e^{- 6 t})

y^{^{'}} - 6 + \frac{y}{( 4 ( x + 250 ) )} = 0, y (0) = 70

\int 2 x sec (4 x^{2}) d x

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