Solving Indefinite Integrals: ∫1/(t*ln(t)) & ∫1/(√(t)*[1-2*√(t)])

Justabeginner
Messages
309
Reaction score
1

Homework Statement


∫1/(t*ln(t)) dt

∫1/(√(t)*[1-2*√(t)]) dt


Homework Equations





The Attempt at a Solution



I used u-substitution for both. For the first equation, my u= ln t, and my final answer was ln|u| + C, or ln(ln(|t|) + C. For the second equation, my u= 1-2*√(t) and my final answer was -ln(u) or -ln(1-2*√(t)) + C.

But both of these are wrong, so I'm not sure what I did wrong. Can I get some guidance on this? Thank you. :)
 
Physics news on Phys.org
Justabeginner said:

Homework Statement


∫1/(t*ln(t)) dt

∫1/(√(t)*[1-2*√(t)]) dt


Homework Equations





The Attempt at a Solution



I used u-substitution for both. For the first equation, my u= ln t, and my final answer was ln|u| + C, or ln(ln(|t|) + C. For the second equation, my u= 1-2*√(t) and my final answer was -ln(u) or -ln(1-2*√(t)) + C.

But both of these are wrong, so I'm not sure what I did wrong. Can I get some guidance on this? Thank you. :)

Both of those are correct as far as integration goes. Must be an error in the answering system.
 
Oh wow, that makes me feel better now. Thank you! :)
 
Thank you! That site helps.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
Back
Top