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

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

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

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

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

Solution steps

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

Expand

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

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

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

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

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

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

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

Simplify

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

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

Add a constant to the solution

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

\int \frac{x ^{3} + 3 x ^{2} + 3 x + 63}{( x ^{2} - 9 ) ^{2}} d x

a x

\int_{0}^{\frac{π}{3}} x sin (9 x) d x

\frac{1}{( s ^{2} + 1 ) ^{2}}

\frac{x ^{2}}{7 + 8 x}

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