how to integrate ( x^2+2x+2)/(x^3 e^x)
- print Print
- list Cite
Expert Answers
calendarEducator since 2011
write5,349 answers
starTop subjects are Math, Science, and Business
You need to use the linearity of integrals such that:
`int (x^2+2x+2)/(x^3 e^x) dx = int (x^2)/(x^3)e^(-x)dx + int (2x)/(x^3)*e^(-x)dx + int 2/x^3 e^(-x)dx`
`int (x^2+2x+2)/(x^3 e^x) dx = int 1/x e^(-x)dx + 2 int 1/x^2 e^(-x)dx + 2 int 1/x^3 e^(-x)dx`
You need to use integration by parts to solve `int 1/x e^(-x)dx` such that:
`int udv = uv - int vdu`
`u = 1/x => du = -1/x^2`
`dv = e^(-x) => v = -e^(-x)`
`int 1/x e^(-x)dx = -e^(-x)/x - int 1/x^2 e^(-x) dx`
`int (x^2+2x+2)/(x^3 e^x) dx = -e^(-x)/x - int 1/x^2 e^(-x) dx + 2 int 1/x^2 e^(-x)dx + 2 int 1/x^3 e^(-x)dx`
`int (x^2+2x+2)/(x^3 e^x) dx = -e^(-x)/x + int 1/x^2 e^(-x)dx + 2 int 1/x^3 e^(-x)dx `
You need to use integration by parts to solve `int 1/x^3 e^(-x)dx` such that:
`int 1/x^3 e^(-x)dx = -1/(2x^2) e^(-x) - (1/2)int 1/x^2 e^(-x)dx `
`u = e^(-x) => du = -e^(-x)`
`dv = 1/x^3 => v = -1/(2x^2)`
`int (x^2+2x+2)/(x^3 e^x) dx = -e^(-x)/x + int 1/x^2 e^(-x)dx -1/(x^2) e^(-x) - int 1/x^2 e^(-x)dx`
Reducing like terms yields:
`int (x^2+2x+2)/(x^3 e^x) dx = -e^(-x)/x -1/(x^2) e^(-x) + c`
`int (x^2+2x+2)/(x^3 e^x) dx = -e^(-x)/x(1 + 1/x) + c`
Hence, evaluating the given integral, using mostly integration by parts, `int (x^2+2x+2)/(x^3 e^x) dx = -e^(-x)/x(1 + 1/x) + c.`
Related Questions
- Calculate the definite integral of f(x)=e^2x from x=0 to x=2.
- 1 Educator Answer
- Solve for X. e^2x - 3e^x + 2=0
- 1 Educator Answer
- How do you find the integral of e^(-x)sin(2x)dx using integration by parts?Here's my work: u =...
- 1 Educator Answer
- `int (x e^(2x))/(1 + 2x)^2 dx` Evaluate the integral
- 1 Educator Answer
- `int (x^3 + 2x^2 + 3x - 2)/(x^2 + 2x + 2)^2 dx` Evaluate the integral
- 1 Educator Answer
Unlock This Answer Now
Start your 48-hour free trial to unlock this answer and thousands more. Enjoy eNotes ad-free and cancel anytime.