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

  • Thread starter a.a
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  • #1
a.a
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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...
 

Answers and Replies

  • #2
statdad
Homework Helper
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If you mean

[tex]
\int e^{x^2+x} (2x+1) \, dx
[/tex]

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

[tex]
\int \frac 1 u \, du
[/tex]

as the next step?
 
  • #3
a.a
127
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because du=e(x2 +x)(2x+1)dx ?

integral e(x2 +x)(2x+1) dx = integral du/e(x2 +x)
=integral 1/u du
?
 
  • #4
statdad
Homework Helper
1,495
35


If

[tex]
u = e^{x^2+x}
[/tex]

then

[tex]
du = e^{x^2+x}(2x+1) dx
[/tex]


which is exactly the form of the original integral.
why is there need for a fraction?
 

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