What is the integral of (x^3+5x^4-7x)/(x^2) ?

The integral of (x^3+5x^4-7x)/(x^2) is (x^2)/2+(5x^3)/3-7ln|x|+C

Popular >

integral of (x^3+5x^4-7x)/(x^2)

\int \frac{x ^{3} + 5 x ^{4} - 7 x}{x ^{2}} d x

\frac{x ^{2}}{2} + \frac{5 x ^{3}}{3} - 7 ln ∣ x ∣ + C

Solution steps

\int \frac{x ^{3} + 5 x ^{4} - 7 x}{x ^{2}} d x

Expand

\frac{x ^{3} + 5 x ^{4} - 7 x}{x ^{2}} : x + 5 x^{2} - \frac{7}{x}

= \int x + 5 x^{2} - \frac{7}{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 d x + \int 5 x^{2} d x - \int \frac{7}{x} d x

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

\int 5 x^{2} d x = \frac{5 x ^{3}}{3}

\int \frac{7}{x} d x = 7 ln ∣ x ∣

= \frac{x ^{2}}{2} + \frac{5 x ^{3}}{3} - 7 ln ∣ x ∣

Add a constant to the solution

= \frac{x ^{2}}{2} + \frac{5 x ^{3}}{3} - 7 ln ∣ x ∣ + C

\int_{1}^{2} 5 x - 1 d x

\int_{- \infty}^{3} \frac{1}{x ^{2} + 9} d x

\frac{\partial}{\partial y} (\frac{6 x}{3 x ^{2} - y ^{3} - 3})

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

x^{7} - 3 x^{4} + 7 x^{3} + 43, at x = - 2

What is the integral of (x^3+5x^4-7x)/(x^2) ?
The integral of (x^3+5x^4-7x)/(x^2) is (x^2)/2+(5x^3)/3-7ln|x|+C