Question

Find the indefinit integral x^3-1/x^2 dx​

Answer 1

The indefinite integral of the function x^3 - 1/x^2 dx is x^4/4 + 1/x + C, where C represents the constant of integration.

Step-by-step explanation:

The student has asked to find the indefinite integral of the function x^3-1/x^2 dx. To find this integral, we would separate the integrand into two parts and integrate each one separately. The integral of x^3 with respect to dx is x^4/4, as we add 1 to the exponent and then divide by the new exponent. The integral of -1/x^2 with respect to dx is 1/x because we use the power rule for integration, which involves adding 1 to the exponent (-2) and then dividing by the new exponent (-1), resulting in a negative sign. Thus, the integral of x^3 would be x^4/4, and for -1/x^2, it would be +1/x. Remember to include the constant of integration, typically denoted as C.

So, the indefinite integral of x^3-1/x^2 dx is x^4/4 + 1/x + C.

Answer 2

`\displaystyle \int {(x^3 - 1)/(x^2)} \, dx = (x^2)/(2) + (1)/(x) + C`

General Formulas and Concepts:

Algebra I

• Exponential Property [Dividing]:
  `\displaystyle (b^m)/(b^n) = b^(m - n)`
• Exponential Property [Rewrite]:
  `\displaystyle b^(-m) = (1)/(b^m)`

Calculus

Integration

• Integrals
• [Indefinite Integrals] integration Constant C

Integration Rule [Reverse Power Rule]:
`\displaystyle \int {x^n} \, dx = (x^(n + 1))/(n + 1) + C`

Integration Property [Multiplied Constant]:
`\displaystyle \int {cf(x)} \, dx = c \int {f(x)} \, dx`

Integration Property [Addition/Subtraction]:
`\displaystyle \int {[f(x) \pm g(x)]} \, dx = \int {f(x)} \, dx \pm \int {g(x)} \, dx`

Step-by-step explanation:

Step 1: Define

Identify

`\displaystyle \int {(x^3 - 1)/(x^2)} \, dx`

Step 2: Integrate

1. [Integrand] Rewrite:
   `\displaystyle \int {(x^3 - 1)/(x^2)} \, dx = \int {\bigg( (x^3)/(x^2) - (1)/(x^2) \bigg)} \, dx`
2. Simplify [Exponential Property - Dividing]:
   `\displaystyle \int {(x^3 - 1)/(x^2)} \, dx = \int {\bigg( x - (1)/(x^2) \bigg)} \, dx`
3. [Integral] Rewrite [Integration Property - Addition/Subtraction]:
   `\displaystyle \int {(x^3 - 1)/(x^2)} \, dx = \int {x} \, dx - \int {(1)/(x^2)} \, dx`
4. [2nd Integral] Rewrite [Exponential Property - Rewrite]:
   `\displaystyle \int {(x^3 - 1)/(x^2)} \, dx = \int {x} \, dx - \int {x^(-2)} \, dx`
5. [Integrals] Reverse Power Rule:
   `\displaystyle \int {(x^3 - 1)/(x^2)} \, dx = (x^2)/(2) - (-x^(-1)) + C`
6. Simplify/Rewrite [Exponential Property - Rewrite]:
   `\displaystyle \int {(x^3 - 1)/(x^2)} \, dx = (x^2)/(2) + (1)/(x) + C`

Topic: AP Calculus AB/BC (Calculus I/I + II)

Unit: Integration