Integral of (x+3)/(x-1) 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:
x−1x+3=1+x−14
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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−14dx=4∫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: 4log(x−1)
The result is: x+4log(x−1)
Method #2
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Rewrite the integrand:
x−1x+3=x−1x+x−13
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Integrate term-by-term:
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Rewrite the integrand:
x−1x=1+x−11
-
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−13dx=3∫x−11dx
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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)
So, the result is: 3log(x−1)
The result is: x+log(x−1)+3log(x−1)
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Add the constant of integration:
x+4log(x−1)+constant
The answer is:
x+4log(x−1)+constant
The answer (Indefinite)
[src]
/
|
| x + 3
| ----- dx = C + x + 4*log(-1 + x)
| x - 1
|
/
∫x−1x+3dx=C+x+4log(x−1)
The graph
−∞−4iπ
=
−∞−4iπ
Use the examples entering the upper and lower limits of integration.