Evaluate in
(cuberootof(x+1)/(x-1)).1/(x+1)dx
help plz
For the life of me, I cannot decipher your expression
I suspect a typo. Wolframalpha indicates that it cannot be done with elementary functions:
http://www.wolframalpha.com/input/?i=integral+%28%28x%2B1%29%2F%28x-1%29%29^%281%2F3%29%2F%28x%2B1%29dx
If I got it wrong, and only the (x+1) is in the cube root, it can then be done, but not easily:
http://www.wolframalpha.com/input/?i=integral+%28x%2B1%29^%281%2F3%29%2F%28%28x-1%29%28x%2B1%29%29dx
Thanks for figuring out what was meant. I just threw up my hands in frustration.
To evaluate the given integral, we can simplify the expression before integrating.
Given: ∫ (cuberoot(x+1) / (x-1)) * (1 / (x+1)) dx
Step 1: Simplify the expression
First, let's simplify the expression inside the integral.
The numerator can be written as (x+1)^(1/3) using the cube root property.
Using the property of exponents, and multiplying the two fractions, we get:
∫ [(x+1)^(1/3) / ((x-1) * (x+1))] dx
Simplifying the denominator further, we get:
∫ (x+1)^(1/3) / (x^2 - 1) dx
Step 2: Split the Fraction
Next, we will split the fraction into two separate fractions to make it easier to integrate.
∫ (x+1)^(1/3) / (x^2 - 1) dx = ∫ (x+1)^(1/3) / [(x-1)(x+1)] dx
Split the denominator:
= ∫ [(x+1)^(1/3) / (x-1)] / (x+1) dx
= ∫ [1 / (x-1)] * [(x+1)^(1/3) / (x+1)] dx
Step 3: Evaluate the integrals separately
Now, we have two separate fractions to integrate.
∫ [1 / (x-1)] dx --> Let's call this integral A
∫ [(x+1)^(1/3) / (x+1)] dx --> Let's call this integral B
Integral A:
This is a basic integral of ln|x-1| + C (where C is the constant of integration).
Integral B:
For this integral, we can simplify it further.
Simplify the expression inside the integral:
∫ [(x+1)^(1/3) / (x+1)] dx
Using a substitution u = x+1, we can rewrite the integral as:
∫ u^(1/3) / u du
Simplify by canceling out the common terms:
∫ u^(-2/3) du
Integrating this, we get:
(3/1) * u^(1/3) + C1
Substituting u back as x+1:
(3/1) * (x+1)^(1/3) + C1
Step 4: Combine the results
Now that we have evaluated both integrals A and B, we can combine them to get the final result.
∫ (x+1)^(1/3) / (x^2 - 1) dx = ln|x-1| + [3/(x+1)^(2/3)] + C2
where C2 represents the constant of integration for this integral.
Therefore, the given integral evaluates to ln|x-1| + [3/(x+1)^(2/3)] + C2.