What is the integral of ((3x^5-6x^4+5x^3)/(x^4)) ?

The integral of ((3x^5-6x^4+5x^3)/(x^4)) is (3x^2)/2-6x+5ln|x|+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of ((3x^5-6x^4+5x^3)/(x^4))

\int (\frac{3 x ^{5} - 6 x ^{4} + 5 x ^{3}}{x ^{4}}) d x

\frac{3 x ^{2}}{2} - 6 x + 5 ln ∣ x ∣ + C

Solution steps

\int \frac{3 x ^{5} - 6 x ^{4} + 5 x ^{3}}{x ^{4}} d x

Simplify

\frac{3 x ^{5} - 6 x ^{4} + 5 x ^{3}}{x ^{4}} : \frac{3 x ^{2} - 6 x + 5}{x}

= \int \frac{3 x ^{2} - 6 x + 5}{x} d x

Expand

\frac{3 x ^{2} - 6 x + 5}{x} : 3 x - 6 + \frac{5}{x}

= \int 3 x - 6 + \frac{5}{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 3 x d x - \int 6 d x + \int \frac{5}{x} d x

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

\int 6 d x = 6 x

\int \frac{5}{x} d x = 5 ln ∣ x ∣

= \frac{3 x ^{2}}{2} - 6 x + 5 ln ∣ x ∣

Add a constant to the solution

= \frac{3 x ^{2}}{2} - 6 x + 5 ln ∣ x ∣ + C

\int 3 x e^{4 x^{2}} d x

\int \frac{x ^{2} + 11 x}{( x + 1 ) ^{2} ( x - 1 )} d x

\int \frac{5}{x + 5} d x

\int 100 e^{4 x} d x

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

What is the integral of ((3x^5-6x^4+5x^3)/(x^4)) ?
The integral of ((3x^5-6x^4+5x^3)/(x^4)) is (3x^2)/2-6x+5ln|x|+C