Integrate ? #(2x+1)e^(x^2+x) dx#

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Answer 1 · 2018-04-03T08:31:07

#int(2x+1)e^(x^2+x) dx#

Let #x^2+x = t#
#(2x+1) dx = dt#

Plugging in these values,

#=>int(2x+1)e^(x^2+x) dx = inte^t dt#

#=> e^t + c#

#=> e^(x^2+x) + c#

Answer 2 · 2018-04-03T08:33:29

#int(2x+1)e^(x^2+x)dx=e^(x^2+x)+C#

.

#int(2x+1)e^(x^2+x)dx#

Let #u=x^2+x, :. du=(2x+1)dx#

Let's substitute:

#inte^udu=e^u+C#

Now, we can substitute back:

#int(2x+1)e^(x^2+x)dx=e^(x^2+x)+C#