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

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

Popular >

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

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

\frac{x ^{3}}{3} + ln x^{2} + x + C

Solution steps

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

Long division

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

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

\int x^{2} d x = \frac{x ^{3}}{3}

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

= \frac{x ^{3}}{3} + ln x^{2} + x

Add a constant to the solution

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

\frac{d}{d x} (\frac{x ^{2} + 5}{x ^{2} - 1})

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

\int_{0}^{1} \frac{61}{4 y - 1} d y

\frac{d y}{d x}, ln (x^{2} - 24 y) = x - y - 4

f (x) = - 2 - 6 x^{2}, (5, - 152)

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