Integral of 2x-3/(x-1)(x+2) dx
The solution
Detail solution
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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:
∫2xdx=2∫xdx
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The integral of xn is n+1xn+1 when n=−1:
∫xdx=2x2
So, the result is: x2
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The integral of a constant times a function is the constant times the integral of the function:
∫(−x−13(x+2))dx=−∫x−13(x+2)dx
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The integral of a constant times a function is the constant times the integral of the function:
∫x−13(x+2)dx=3∫x−1x+2dx
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There are multiple ways to do this integral.
Method #1
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Rewrite the integrand:
x−1x+2=1+x−13
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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−13dx=3∫x−11dx
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Let u=x−1.
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−1)
So, the result is: 3log(x−1)
The result is: x+3log(x−1)
Method #2
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Rewrite the integrand:
x−1x+2=x−1x+x−12
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Integrate term-by-term:
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Rewrite the integrand:
x−1x=1+x−11
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Integrate term-by-term:
-
The integral of a constant is the constant times the variable of integration:
∫1dx=x
-
Let u=x−1.
Then let du=dx and substitute du:
∫u1du
-
The integral of u1 is log(u).
Now substitute u back in:
log(x−1)
The result is: x+log(x−1)
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The integral of a constant times a function is the constant times the integral of the function:
∫x−12dx=2∫x−11dx
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Let u=x−1.
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−1)
So, the result is: 2log(x−1)
The result is: x+log(x−1)+2log(x−1)
So, the result is: 3x+9log(x−1)
So, the result is: −3x−9log(x−1)
The result is: x2−3x−9log(x−1)
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Add the constant of integration:
x2−3x−9log(x−1)+constant
The answer is:
x2−3x−9log(x−1)+constant
The answer (Indefinite)
[src]
/
|
| / 3 \ 2
| |2*x - -----*(x + 2)| dx = C + x - 9*log(-1 + x) - 3*x
| \ x - 1 /
|
/
∫(2x−x−13(x+2))dx=C+x2−3x−9log(x−1)
The graph
∞+9iπ
=
∞+9iπ
Use the examples entering the upper and lower limits of integration.