# Solving simple integral [closed]

How to solve next integral: $$\int\frac{x^3+4x^2-5}{x^2}dx$$ I am using power rule for top part wich produce this: $$\int\frac{\frac{x^2}{2}+4x-5x}{x^2}dx$$ Does my calculation right? How can I continue from here? Please describe all steps and rules used for solving this integral.

## closed as off-topic by Jack D'Aurizio, Namaste, Arnaldo, Leucippus, Daniel W. FarlowJun 7 '17 at 0:49

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• It is obviously wrong. Otherwise $\frac{x^2}{2}=\int x\,dx = \int\frac{x^2}{x}\,dx = \frac{\frac{x^3}{3}}{x}=\frac{x^2}{3}$, according to your logic. – Jack D'Aurizio Jun 6 '17 at 20:18
• Which power rule? – Qwerty Jun 6 '17 at 20:18
• I can not see any rule. $x^3\to \frac12x^2$, $x^2\to x$ and $1\to x$? That can not be right. – Mundron Schmidt Jun 6 '17 at 20:21
• remember to simplify the exponents first!! – Saketh Malyala Jun 6 '17 at 20:32

What you already have is incorrect. Try this:

$$\int \frac{x^3+4x^2-5}{x^2}dx = \int \frac{x^3}{x^2} + \frac{4x^2}{x^2} -\frac{5}{x^2} dx$$

$$=\int (x) dx + \int 4 dx + \int \frac{5}{x^2} dx$$

$$=\frac{x^2}{2} + 4x + \frac{5}{x} + C$$

Overview of Integration rules:

$$\int x^n \hspace{2 mm}dx= \frac{x^{n+1}}{n+1} + C$$

• How $\frac{x^3}{x^2}$ becomes $\frac{x^2}{2}$? and how $-\frac{5}{x^2}$ becomes ${5}{x}$? – IntoTheDeep Jun 6 '17 at 20:25
• $$\int x^n = \frac{x^{n+1}}{n+1} + C$$ Check this out as a reference: mathsisfun.com/calculus/integration-rules.html – Dashi Jun 6 '17 at 20:30
• Does this reference apply for $-\frac{5}{x^2}$? – IntoTheDeep Jun 6 '17 at 20:34
• Yes: $$\int \frac{-5}{x^2} dx= -5 \int x^{-2} dx = -5 \left( \frac{x^{-2+1}}{-2+1} \right)+C = -5 (-x^{-1}) + C = \frac{5}{x} + C$$ – Dashi Jun 6 '17 at 20:36
• $$\int \frac{x^3}{x^2} dx = \int x^1 \hspace{2 mm} dx = \frac{x^{1+1}}{1+1} + C = \frac{x^2}{2} + C$$ – Dashi Jun 6 '17 at 20:51

Hint:

$$\int \frac{x^3+4x^2-5}{x^2}dx=\int x+4-5x^{-2}dx$$

What you want to use is the linearity of the integral: $$\int \frac{x^3 + 4x^2 -5}{x^2} dx = \int \frac{x^3}{x^2} dx + \int \frac{ 4x^2}{x^2} dx - \int \frac{5}{x^2} dx.$$Can you continue from here on out?
\begin{align}\int\frac{x^3+4x^2-5}{x^2}dx=&\int{\left(x+4-{5\over x^2}\right)}dx\\ =&\int{xdx}+4\int{dx}-5\int\frac{dx}{x^2}\\ =&{x^2\over 2}+4x+{5\over x}+C\end{align}