Is using u-substitution the right approach for integrating e(x^2 + x)(2x+1) dx?

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Homework Help Overview

The problem involves integrating the expression e^(x^2 + x)(2x + 1) dx, which falls under the subject area of calculus, specifically integration techniques.

Discussion Character

  • Exploratory, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to use u-substitution by letting u = e^(x^2 + x) and derives du accordingly. Some participants question the transition to the integral of 1/u du, seeking clarification on the steps involved.

Discussion Status

The discussion is ongoing, with participants exploring different substitution methods and questioning the necessity of certain steps in the integration process. There is no explicit consensus yet, but participants are engaging with each other's reasoning.

Contextual Notes

Some participants are addressing potential misunderstandings regarding the application of u-substitution and the form of the integral after substitution. The original poster mentions not obtaining the same answer as the professor, indicating a possible discrepancy in understanding or execution.

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Homework Statement



integrate: e(x^2 +x)(2x+1) dx


The Attempt at a Solution



let u= e(x^2 +x)
du=e(x^2 +x)(2x+1)dx

integral e(x^2 +x)(2x+1) dx = integral 1/u du

am I on the right track? i didnt get the same answer as the prof...
 
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If you mean

<br /> \int e^{x^2+x} (2x+1) \, dx<br />

then your substitution is one of at least two that will work, but how do you obtain

<br /> \int \frac 1 u \, du <br />

as the next step?
 


because du=e(x2 +x)(2x+1)dx ?

integral e(x2 +x)(2x+1) dx = integral du/e(x2 +x)
=integral 1/u du
?
 


If

<br /> u = e^{x^2+x}<br />

then

<br /> du = e^{x^2+x}(2x+1) dx<br />which is exactly the form of the original integral.
why is there need for a fraction?
 

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