Integral of (3x+2)/(x-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=3x.
Then let du=3dx and substitute du:
∫u−9u+2du
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Let u=u−9.
Then let du=du and substitute du:
∫uu+11du
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Rewrite the integrand:
uu+11=1+u11
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Integrate term-by-term:
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The integral of a constant is the constant times the variable of integration:
∫1du=u
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The integral of a constant times a function is the constant times the integral of the function:
∫u11du=11∫u1du
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The integral of u1 is log(u).
So, the result is: 11log(u)
The result is: u+11log(u)
Now substitute u back in:
u+11log(u−9)−9
Now substitute u back in:
3x+11log(3x−9)−9
Method #2
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Rewrite the integrand:
x−33x+2=3+x−311
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Integrate term-by-term:
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The integral of a constant is the constant times the variable of integration:
∫3dx=3x
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The integral of a constant times a function is the constant times the integral of the function:
∫x−311dx=11∫x−31dx
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Let u=x−3.
Then let du=dx and substitute du:
∫u1du
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The integral of u1 is log(u).
Now substitute u back in:
log(x−3)
So, the result is: 11log(x−3)
The result is: 3x+11log(x−3)
Method #3
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Rewrite the integrand:
x−33x+2=x−33x+x−32
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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:
∫x−33xdx=3∫x−3xdx
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Rewrite the integrand:
x−3x=1+x−33
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Integrate term-by-term:
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The integral of a constant is the constant times the variable of integration:
∫1dx=x
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The integral of a constant times a function is the constant times the integral of the function:
∫x−33dx=3∫x−31dx
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Let u=x−3.
Then let du=dx and substitute du:
∫u1du
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The integral of u1 is log(u).
Now substitute u back in:
log(x−3)
So, the result is: 3log(x−3)
The result is: x+3log(x−3)
So, the result is: 3x+9log(x−3)
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The integral of a constant times a function is the constant times the integral of the function:
∫x−32dx=2∫x−31dx
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Let u=x−3.
Then let du=dx and substitute du:
∫u1du
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The integral of u1 is log(u).
Now substitute u back in:
log(x−3)
So, the result is: 2log(x−3)
The result is: 3x+9log(x−3)+2log(x−3)
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Add the constant of integration:
3x+11log(3x−9)−9+constant
The answer is:
3x+11log(3x−9)−9+constant
The answer (Indefinite)
[src]
/
|
| 3*x + 2
| ------- dx = -9 + C + 3*x + 11*log(-9 + 3*x)
| x - 3
|
/
∫x−33x+2dx=C+3x+11log(3x−9)−9
The graph
3 - 11*log(3) + 11*log(2)
−11log(3)+3+11log(2)
=
3 - 11*log(3) + 11*log(2)
−11log(3)+3+11log(2)
3 - 11*log(3) + 11*log(2)
Use the examples entering the upper and lower limits of integration.