How do you integrate the following questions?

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In summary, to integrate variables into an equation, one must use the basic rules of integration such as the power rule, product rule, quotient rule, or chain rule. The difference between indefinite and definite integration lies in the resulting solution, with indefinite integration producing an equation with a constant and definite integration giving a numerical value. The bounds of a definite integral are determined by the limits of integration, usually represented by the beginning and end values of the integration interval. The purpose of integration is to find the area under a curve, solve differential equations, and find the average value of a function. The fundamental theorem of calculus allows for the use of shortcuts in evaluating integrals by finding the antiderivative of a function and evaluating it at the upper and lower
  • #1
jkh4
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How do you integrate the following questions?

1) x/((3-x^4)^(1/2))*dx

2) (1+e^x)/(1-e^x)*dx

3) (xlnx)/((x^2-1)^(1/2))*dx

thank you so much!
 
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  • #2
1. Set [tex] x^{2} = \sqrt{3}\sin \theta [/tex]2. Divide it and get [tex] \int 1 + \frac{2e^{x}}{1-e^{x}} dx [/tex]
3. Use integration by parts.
 
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  • #3


To integrate these questions, we can use various integration techniques such as substitution, integration by parts, and trigonometric substitution. Here are some possible approaches for each question:

1) For the first question, we can use the substitution method by letting u = 3-x^4. Then, du = -4x^3 dx and x^4 = 3-u. Substituting these into the integral, we get:

∫ x/((3-x^4)^(1/2))*dx = ∫ -1/(2√u) du = -√u + C = -√(3-x^4) + C

2) The second question can be solved using the substitution method as well. Let u = e^x, then du = e^x dx. The integral becomes:

∫ (1+e^x)/(1-e^x)*dx = ∫ (1+u)/(1-u) du = ∫ (2/(1-u)) - 1 du = 2ln|1-u| - u + C = 2ln|1-e^x| - e^x + C

3) For the third question, we can use integration by parts with u = lnx and dv = (x^2-1)^(-1/2)dx. Then, du = 1/x dx and v = 2√(x^2-1). The integral becomes:

∫ (xlnx)/((x^2-1)^(1/2))*dx = 2√(x^2-1)lnx - ∫ 2√(x^2-1)/x dx

We can then use the substitution method by letting u = x^2-1, du = 2x dx, and x = √(u+1). The integral becomes:

∫ 2√(x^2-1)/x dx = ∫ 2√(u+1)/(√(u+1)) du = 2∫ √(u+1) du = (4/3)(u+1)^(3/2) + C = (4/3)(x^2)^3/2 + C = (4/3)x^3 + C

Putting everything together, we get the final answer as:

∫ (xlnx)/((x
 

Related to How do you integrate the following questions?

1. How do you integrate variables into an equation?

To integrate variables into an equation, you need to follow the basic rules of integration. You can use the power rule, product rule, quotient rule, or chain rule depending on the type of equation. Make sure to properly substitute in the variable's value and simplify the equation before integrating.

2. What is the difference between indefinite and definite integration?

Indefinite integration is the process of finding a general solution to an equation, while definite integration involves finding the exact value of a definite integral over a specific interval. Indefinite integration results in an equation with a constant as the integration constant, while definite integration gives a numerical value.

3. How do you determine the bounds of a definite integral?

The bounds of a definite integral are determined by the limits of integration, which are usually represented by the values of the independent variable at the beginning and end of the integration interval. The lower bound is typically denoted by a and the upper bound by b.

4. What is the purpose of integration in mathematics?

The purpose of integration is to find the area under a curve or the accumulation of a quantity over a specific interval. It is also used to solve differential equations and to find the average value of a function.

5. How do you use the fundamental theorem of calculus to integrate a function?

The fundamental theorem of calculus states that the definite integral of a function can be calculated by finding the antiderivative of the function and evaluating it at the upper and lower bounds of the integration interval. This theorem allows for the use of shortcuts in evaluating integrals and is a fundamental concept in calculus.

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