Question

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

Answer 1

Hint: 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:
\[\int {dx = x + c} \],
\[\int {adx = ax + c} \],
\[\int {{x^n}dx = \dfrac{{{x^{n + 1}}}}{{n + 1}}} + c\],
\[\int {\dfrac{1}{x}dx} = \ln x + c\],
\[\int {{e^x}dx = {e^x} + c} \],
\[\int {{a^x}dx = \dfrac{{{a^x}}}{{\ln a}} + c} \],
\[\int {\sin xdx = - \cos x + c} \],
\[\int {\cos xdx = \sin x + c} \],
\[\int {{{\sec }^2}xdx = \tan x + c} \],
\[\int {{{\csc }^2}xdx = - \cot x + C} \],
\[\int {\sec x \tan xdx = \sec x + C} \],
\[\int {\csc x\cot xdx = - \csc x + C} \],
\[\int {\dfrac{1}{{\sqrt {1 - {x^2}} }}dx = {{\sin }^{ - 1}}x + c} \],
\[\int {\dfrac{1}{{1 + {x^2}}}dx = {{\tan }^{ - 1}}x + c} \].