Question

How do you find the indefinite integral of $\int {{r^2} - 2r + \dfrac{1}{r}dr}$?

Answer 1

In solving the given question, we have to integrate the given function, first distribute the integration to all terms in the given function, then by using integration formulas such as $\int {{x^n}dx = \dfrac{{{x^{n + 1}}}}{{n + 1}}} + c$ and $\int {\dfrac{1}{x}dx} = \ln x + c$ wherever required, and then further simplification we will get the required result.

Complete step by step solution:
Indefinite integral refers to an integral that does not have any upper and lower limit.
Given function is $\int {{r^2} - 2r + \dfrac{1}{r}dr}$,
Now distribute the integral to all the terms of the function, we get,
$\Rightarrow \int {{r^2} - 2r + \dfrac{1}{r}dr} = \int {{r^2}dr - \int {2r} dr + \int {\dfrac{1}{r}} dr}$,
Again rewrite the expression by taking the constant out from the second term we get,
$\Rightarrow \int {{r^2} - 2r + \dfrac{1}{r}dr} = \int {{r^2}dr - 2\int r dr + \int {\dfrac{1}{r}} dr}$,
Now integrate each part using integration formulas such as , $\int {{x^n}dx = \dfrac{{{x^{n + 1}}}}{{n + 1}}} + c$and$\int {\dfrac{1}{x}dx} = \ln x + c$, we get,
For the first term i.e., ${r^2}$, $n = 2$ and $x = r$, and for the second term i.e.,$n = 1$ and $x = r$, now substituting the values in the formulas we get,
$\Rightarrow \int {{r^2} - 2r + \dfrac{1}{r}dr} = \dfrac{{{r^{2 + 1}}}}{{2 + 1}} - 2\dfrac{{{r^{1 + 1}}}}{{1 + 1}} + \ln \left| r \right|$,
Now simplifying powers and denominators, we get,
$\Rightarrow \int {{r^2} - 2r + \dfrac{1}{r}dr} = \dfrac{{{r^3}}}{3} - 2\dfrac{{{r^2}}}{2} + \ln \left| r \right|$,
Now further simplification by eliminating the like terms, we get,
$\Rightarrow \int {{r^2} - 2r + \dfrac{1}{r}dr} = \dfrac{{{r^3}}}{3} - {r^2} + \ln \left| r \right|$,
So, the integral for the given function is $\dfrac{{{r^3}}}{3} - {r^2} + \ln \left| r \right| + C$.

Final Answer:
$\therefore$ The integral for the given function $\int {{r^2} - 2r + \dfrac{1}{r}dr}$ will be equal to $\dfrac{{{r^3}}}{3} - {r^2} + \ln \left| r \right| + C$.

Note:
An indefinite is an integral that does not contain the upper and lower limit. Indefinite integral is also known as anti-derivative or Prime integral. Indefinite integral of a function f is generally a differentiable function F whose derivative is equal to the original function f. It is represented as,
$\int {f\left( x \right)dx} = F\left( x \right) + C$,
Some of the important formulas that we use while solving integration problems are given below:
$\int {dx = x + c}$,
$\int {adx = ax + c}$,
$\int {{x^n}dx = \dfrac{{{x^{n + 1}}}}{{n + 1}}} + c$,
$\int {\dfrac{1}{x}dx} = \ln x + c$,
$\int {{e^x}dx = {e^x} + c}$,
$\int {{a^x}dx = \dfrac{{{a^x}}}{{\ln a}} + c}$,
$\int {\sin xdx = - \cos x + c}$,
$\int {\cos xdx = \sin x + c}$,
$\int {{{\sec }^2}xdx = \tan x + c}$,
$\int {{{\csc }^2}xdx = - \cot x + C}$,
$\int {\sec x \tan xdx = \sec x + C}$,
$\int {\csc x\cot xdx = - \csc x + C}$,
$\int {\dfrac{1}{{\sqrt {1 - {x^2}} }}dx = {{\sin }^{ - 1}}x + c}$,
$\int {\dfrac{1}{{1 + {x^2}}}dx = {{\tan }^{ - 1}}x + c}$.