Integral of (2x+3)^3 dx
The solution
Detail solution
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There are multiple ways to do this integral.
Method #1
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Let u=2x+3.
Then let du=2dx and substitute 2du:
∫2u3du
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The integral of a constant times a function is the constant times the integral of the function:
∫u3du=2∫u3du
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The integral of un is n+1un+1 when n=−1:
∫u3du=4u4
So, the result is: 8u4
Now substitute u back in:
8(2x+3)4
Method #2
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Rewrite the integrand:
(2x+3)3=8x3+36x2+54x+27
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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:
∫8x3dx=8∫x3dx
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The integral of xn is n+1xn+1 when n=−1:
∫x3dx=4x4
So, the result is: 2x4
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The integral of a constant times a function is the constant times the integral of the function:
∫36x2dx=36∫x2dx
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The integral of xn is n+1xn+1 when n=−1:
∫x2dx=3x3
So, the result is: 12x3
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The integral of a constant times a function is the constant times the integral of the function:
∫54xdx=54∫xdx
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The integral of xn is n+1xn+1 when n=−1:
∫xdx=2x2
So, the result is: 27x2
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The integral of a constant is the constant times the variable of integration:
∫27dx=27x
The result is: 2x4+12x3+27x2+27x
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Now simplify:
8(2x+3)4
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Add the constant of integration:
8(2x+3)4+constant
The answer is:
8(2x+3)4+constant
The answer (Indefinite)
[src]
/
| 4
| 3 (2*x + 3)
| (2*x + 3) dx = C + ----------
| 8
/
∫(2x+3)3dx=C+8(2x+3)4
The graph
Use the examples entering the upper and lower limits of integration.