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

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

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

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

\frac{1}{2} x^{2} - 2 x + 2 ln ∣ x + 1 ∣ + \frac{2}{x + 1} + ln ∣ x ∣ + \frac{1}{2} - 3 + C

Solution steps

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

Expand

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

\int \frac{x ^{4}}{x ^{3} + 2 x ^{2} + x} d x = \frac{1}{2} (x + 1)^{2} - 3 (x + 1) + 3 ln ∣ x + 1 ∣ + \frac{1}{x + 1}

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

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

Simplify

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

= \frac{1}{2} x^{2} - 2 x + 2 ln ∣ x + 1 ∣ + \frac{2}{x + 1} + ln ∣ x ∣ + \frac{1}{2} - 3

Add a constant to the solution

= \frac{1}{2} x^{2} - 2 x + 2 ln ∣ x + 1 ∣ + \frac{2}{x + 1} + ln ∣ x ∣ + \frac{1}{2} - 3 + C

\frac{d}{d x} (- \frac{6}{( x + 1 ) ^{4}})

x \to \infty lim (x \cdot \frac{x}{x})

\frac{d}{d t} (\frac{3 t ^{2} - 3}{2 t})

\frac{d}{d x} (x^{2} + 2 x + 6)

\int_{0}^{1} \frac{27}{4 y - 1} d y

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