Integral of (3x+1)e^x dx
The solution
Detail solution
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There are multiple ways to do this integral.
Method #1
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Rewrite the integrand:
(3x+1)ex=3xex+ex
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Integrate term-by-term:
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The integral of a constant times a function is the constant times the integral of the function:
∫3xexdx=3∫xexdx
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Use integration by parts:
∫udv=uv−∫vdu
Let u(x)=x and let dv(x)=ex.
Then du(x)=1.
To find v(x):
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The integral of the exponential function is itself.
∫exdx=ex
Now evaluate the sub-integral.
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The integral of the exponential function is itself.
∫exdx=ex
So, the result is: 3xex−3ex
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The integral of the exponential function is itself.
∫exdx=ex
The result is: 3xex−2ex
Method #2
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Use integration by parts:
∫udv=uv−∫vdu
Let u(x)=3x+1 and let dv(x)=ex.
Then du(x)=3.
To find v(x):
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The integral of the exponential function is itself.
∫exdx=ex
Now evaluate the sub-integral.
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The integral of a constant times a function is the constant times the integral of the function:
∫3exdx=3∫exdx
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The integral of the exponential function is itself.
∫exdx=ex
So, the result is: 3ex
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Now simplify:
(3x−2)ex
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Add the constant of integration:
(3x−2)ex+constant
The answer is:
(3x−2)ex+constant
The answer (Indefinite)
[src]
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| x x x
| (3*x + 1)*e dx = C - 2*e + 3*x*e
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∫(3x+1)exdx=C+3xex−2ex
The graph
Use the examples entering the upper and lower limits of integration.