Integrate dx/(1-x): Solve (2/3a)(1-(1-aW)^(3/2))

Thread starter cycling4life
Start date Jan 31, 2012
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Integration

In summary, the integral of dx/(1-x) is -ln|1-x| + C, where C is the constant of integration. To solve (2/3a)(1-(1-aW)^(3/2)), you can expand the expression using the binomial theorem and then integrate using the power rule for integration. The steps for integrating dx/(1-x) are rewriting the expression, using substitution, integrating using the power rule, and adding the constant of integration. The domain of the function (2/3a)(1-(1-aW)^(3/2)) is generally all real values of a and W, except when (1-aW)^(3/2) results in a negative number. The

Jan 31, 2012

cycling4life

In part of a derivation they have, Integrate dx/(1-x) = (1-aW)^(1/2) dW

I get -ln(1-x) = (-2/3a)*(1-aW)^(3/2) but they say it's ln(1/(1-X)) = (2/3a)((1-(1-AW)^(3/2))

Can anyone tell me how they get that extra "1-"?

Last edited: Jan 31, 2012

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Feb 1, 2012

JJacquelin

-ln(1-x) = (-2/3a)*(1-aW)^(3/2) + Contant

1. What is the integral of dx/(1-x)?

The integral of dx/(1-x) is -ln|1-x| + C, where C is the constant of integration.

2. How do you solve (2/3a)(1-(1-aW)^(3/2))?

To solve (2/3a)(1-(1-aW)^(3/2)), you can expand the expression using the binomial theorem and then integrate using the power rule for integration. The final result will depend on the value of a and W.

3. What are the steps for integrating dx/(1-x)?

The steps for integrating dx/(1-x) are as follows: 1) Rewrite the expression as dx/(-1)(x-1); 2) Use the substitution u = x-1; 3) Substitute u back into the expression; 4) Integrate using the power rule for integration; 5) Add the constant of integration.

4. What is the domain of the function (2/3a)(1-(1-aW)^(3/2))?

The domain of the function (2/3a)(1-(1-aW)^(3/2)) depends on the value of a and W. Generally, the function is defined for all real values of a and W, except for when the expression (1-aW)^(3/2) results in a negative number, in which case the function is undefined.