How do you evaluate the definite integral #int (2x)/(1+x^2)# from #[0,1]#?
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To evaluate the definite integral (\int_{0}^{1} \frac{2x}{1+x^2} , dx), you can use the substitution method.
Let (u = 1 + x^2), then (du = 2x , dx).
Substitute these into the integral:
[\int \frac{2x}{1+x^2} , dx = \int \frac{1}{u} , du]
Now, the integral becomes:
[\int \frac{1}{u} , du = \ln|u| + C]
Substitute back (u = 1 + x^2):
[\ln|1 + x^2| + C]
Evaluate the integral from 0 to 1:
[\ln|1 + 1^2| - \ln|1 + 0^2|]
[= \ln(2) - \ln(1)]
[= \ln(2)]
So, the value of the definite integral (\int_{0}^{1} \frac{2x}{1+x^2} , dx) is (\ln(2)).
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To evaluate the definite integral ( \int_{0}^{1} \frac{2x}{1 + x^2} , dx ), you can use the technique of substitution.
Let ( u = 1 + x^2 ). Then, ( du = 2x , dx ).
Now, rewrite the integral in terms of ( u ):
[ \int_{0}^{1} \frac{2x}{1 + x^2} , dx = \int_{u(0)}^{u(1)} \frac{1}{u} , du ]
The limits of integration also change accordingly:
When ( x = 0 ), ( u = 1 + 0^2 = 1 ).
When ( x = 1 ), ( u = 1 + 1^2 = 2 ).
Now, evaluate the integral with respect to ( u ):
[ \int_{0}^{1} \frac{2x}{1 + x^2} , dx = \left[ \ln|u| \right]_{1}^{2} ]
[ = \ln|2| - \ln|1| ]
[ = \ln|2| - 0 ]
[ = \ln(2) ]
So, the value of the definite integral ( \int_{0}^{1} \frac{2x}{1 + x^2} , dx ) is ( \ln(2) ).
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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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