4
$\begingroup$

I found this picture on Twitter and I'm curious is it solvable or not. For me, a first-year student with little math knowledge, it is just looks like a mess.

enter image description here

Solve for the password : $$ \frac{ \int u (u^2 +5)^{1/2}~du - 3 \int u (u^2 -5)^{-1/2}~ }{\displaystyle\int \frac{u ((u^2 +5)-3)}{\sqrt{u^2 +5}}~du}$$

$\endgroup$
2
  • $\begingroup$ Did you try inputting this formula into Wolfram Alpha? Also attempt substitution on the integrals. $\endgroup$ Oct 8 '19 at 8:04
  • 2
    $\begingroup$ I think the expression is supposed to be $$ \frac{ \int u (u^2 +5)^{1/2}~du - 3 \int u (u^2 -5)^{-1/2}~du }{\int \frac{u ((u^2 +5)-3)}{\sqrt{u^2 +5}}~du} $$ $\endgroup$
    – Matti P.
    Oct 8 '19 at 8:06
4
$\begingroup$

No need to know much of calculus,

$$\frac{u((u^2+5)-3)}{\sqrt{u^2+5}}=u(u^2+5)^{1/2}-3u(u^2+5)^{-1/2}.$$

My bet is that the "Pasword" is "one" or "1", assuming a typo ($-5$ instead of $+5$; also missing $du$), and assuming that the integration constants can be ignored.

$\endgroup$
0
$\begingroup$

It's ridiculous. Numerator and denominator are same, if we more generalise the integration.. So,ans is 1

$\endgroup$
3
  • $\begingroup$ Isn't your answer the same as what Yves has written? $\endgroup$
    – Toby Mak
    Oct 8 '19 at 9:36
  • $\begingroup$ Problem lies in the constant of integration $\endgroup$
    – wilsonw
    Oct 8 '19 at 10:08
  • $\begingroup$ They are quasi the same... $\endgroup$
    – user65203
    Oct 8 '19 at 10:30

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.