Integration by parts involving exponentials and logarithms

In summary, using integration by parts with the given limits, the integral of (1/x^2)(lnx) dx can be solved by setting u=x^-2 and dv/dx=lnx, resulting in [x^-2.x(lnx-1)] - the integral of (-2x^-3.x(lnx-1)). This can then be simplified to x^-2(lnx) - x^-2 + C.
  • #1
xllx
33
0

Homework Statement



Using integration by parts, integrate:
(1/x^2)(lnx) dx with the limits e and 1


Homework Equations



[uv]to the limits a b - the integral of (v)(du/dx) dx

(sorry, don't know how to write out equations properly on a computer)

The Attempt at a Solution



I've got u=x^-2 so du/dx= -2x^-3
dv/dx=lnx so v= x(lnx-1)

So putting this into the equation above:
[x^-2.x(lnx-1)] - the integral of (-2x^-3.x(lnx-1))

Is this right so far?
If so how do I integrate the last part? Do I do it sepeartley or by parts again?

Many Thanks, any help at all would be greatly appreciated.
 
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  • #2
Hi xllx! :smile:

(have an integral: ∫ and try using the X2 tag just above the Reply box :wink:)
xllx said:
Using integration by parts, integrate:
(1/x^2)(lnx) dx with the limits e and 1

I've got u=x^-2 so du/dx= -2x^-3
dv/dx=lnx so v= x(lnx-1)

So putting this into the equation above:
[x^-2.x(lnx-1)] - the integral of (-2x^-3.x(lnx-1))

eugh! :yuck:

go the other way …

integrate x-2 ! :smile:
 
  • #3
Thankyou. Redid it and came out with a reasonal answer.
 

1. How do you integrate a product of an exponential and a logarithm using integration by parts?

To integrate a product of an exponential function and a logarithmic function using integration by parts, you would need to choose one of the functions as the "u" term and the other as the "dv" term. Then, you would use the formula: ∫u(dv) = uv - ∫v(du) to solve the integral. You may need to use the product rule or chain rule to find the derivatives of the "u" and "v" terms.

2. Can integration by parts be used to evaluate definite integrals involving exponentials and logarithms?

Yes, integration by parts can be used to evaluate definite integrals involving exponentials and logarithms. Just like with indefinite integrals, you would choose "u" and "dv" terms and use the formula: ∫u(dv) = uv - ∫v(du) to solve the integral. However, you would also need to apply the limits of integration to the final answer.

3. Are there any special cases when using integration by parts for exponentials and logarithms?

There are a few special cases when using integration by parts for exponentials and logarithms. One is when the exponent of the exponential function is a constant, in which case you can use the integration by parts formula directly. Another is when the logarithmic function is ln(x), in which case the "u" term should be chosen as ln(x) and the "dv" term should be chosen as 1/x.

4. How do you choose which function to use as the "u" term and the "dv" term in integration by parts for exponentials and logarithms?

When choosing which function to use as the "u" term and the "dv" term in integration by parts for exponentials and logarithms, it is generally helpful to choose the function that becomes simpler when differentiated. This will make the integration by parts process easier and more efficient.

5. Can integration by parts be used for more than two functions?

Yes, integration by parts can be used for more than two functions. This is known as the "generalized" integration by parts formula and involves using the product rule multiple times. However, this method can become quite complex and it is usually more efficient to use other integration techniques for integrals involving more than two functions.

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