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

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

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

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

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

Solution steps

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

Factor

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

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

Take the partial fraction of

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

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

\int x d x = \frac{x ^{2}}{2}

\int 1 d x = x

\int \frac{1}{x} d x = ln ∣ x ∣

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

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

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

Add a constant to the solution

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

\frac{\partial}{\partial y} (tan (1 + 2 x^{2} y^{4} z^{2}))

f (x) = 3 sin (x) + ln (5 x)

\int_{- 2}^{2} 4 - x^{2} d x

x \to 2 lim (e^{\frac{1}{2 - x}})

y = 2 + \frac{1}{x}, (2, 2.5)

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