Dtjytyk

Is the destination of a commercial flight important for the pilot?

Large drywall patch supports

How long to clear the 'suck zone' of a turbofan after start is initiated?

What Brexit proposals are on the table in the indicative votes on the 27th of March 2019?

Term for the "extreme-extension" version of a straw man fallacy?

Lay out the Carpet

What is the difference between "behavior" and "behaviour"?

Is there a good way to store credentials outside of a password manager?

Did the DC-9 ever use RATO in revenue service?

Failed to fetch jessie backports repository

Pre-amplifier input protection

Opposite of a diet

Do the temporary hit points from the Battlerager barbarian's Reckless Abandon stack if I make multiple attacks on my turn?

Did Dumbledore lie to Harry about how long he had James Potter's invisibility cloak when he was examining it? If so, why?

How does buying out courses with grant money work?

Was Spock the First Vulcan in Starfleet?

What is paid subscription needed for in Mortal Kombat 11?

How to be diplomatic in refusing to write code that breaches the privacy of our users

How does it work when somebody invests in my business?

What is the intuitive meaning of having a linear relationship between the logs of two variables?

How can I get through very long and very dry, but also very useful technical documents when learning a new tool?

Roman Numeral Treatment of Suspensions

How can I kill an app using Terminal?

You cannot touch me, but I can touch you, who am I?

Find $int frac{x^nln x}{(x^{n+1}+1)^n}dx$, [on hold]

How find this integral $intfrac{x^2+2x+1+(3x+1)sqrt{x+ln{x}}}{xsqrt{x+ln{x}}(x+sqrt{x+ln{x}})}dx$Evaluating: $int 3xsinleft(frac x4right) , dx$.Compute $intfrac{dx}{sqrt{tan x}}$Integrate $intfrac{dx}{(x^2+16)^3}$Computing the Integral $int frac{sqrt{x}}{x^2+x} dx$Integrating $intfrac{x^2-1}{(x^2+1)sqrt{x^4+1}},dx$A problem in calculating integralFinding the integral $int sqrt{1-frac{3}{x}+frac{1}{x^2}} , dx$How can we find a closed form solution of $int(x+a)^m(x+b)^n~dx$?Find the value of $int _{0} ^ {infty} f(x+frac1x)frac{ln x}{x} dx$

0

1

$begingroup$

Find $displaystyleint dfrac{x^nln x}{(x^{n+1}+1)^n}dx$, where $ninmathbb{N}$, $xin(0,infty)$

I think I saw this on mathstack, but I can't find it. Anyway, for this problem, my idea was to consider $x^{n+1}=t$, but that $ln $ put a stop to any valuable development.

calculus integration indefinite-integrals

share|cite|improve this question

edited Mar 15 at 18:48

Mark Viola

134k1278176

asked Mar 15 at 17:08

user651754

$endgroup$

put on hold as off-topic by RRL, mrtaurho, Cesareo, José Carlos Santos, Parcly Taxel yesterday

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – RRL, mrtaurho, Cesareo, José Carlos Santos, Parcly Taxel

If this question can be reworded to fit the rules in the help center, please edit the question.

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Maybe you saw it without the bounds like it was $int_0^infty$. Because this way it doesn't look charming.
  $endgroup$
  – Zacky
  Mar 15 at 17:18
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  It's without the bounds in original. I saw it correctly :))))
  $endgroup$
  – user651754
  Mar 15 at 17:22
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @user651754 Can you provide citations for where you saw it?
  $endgroup$
  – tatan
  Mar 15 at 17:22
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  It's a romanian magazine, called ,,Gazeta Matematica". It's from the febraury issue, 2018.
  $endgroup$
  – user651754
  Mar 15 at 17:24
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Zacky It's not a very difficult indefinite integral. See my post.
  $endgroup$
  – Mark Viola
  Mar 15 at 18:47
  

  
  
  
  

add a comment | 

0

1

$begingroup$

Find $displaystyleint dfrac{x^nln x}{(x^{n+1}+1)^n}dx$, where $ninmathbb{N}$, $xin(0,infty)$

I think I saw this on mathstack, but I can't find it. Anyway, for this problem, my idea was to consider $x^{n+1}=t$, but that $ln $ put a stop to any valuable development.

calculus integration indefinite-integrals

share|cite|improve this question

edited Mar 15 at 18:48

Mark Viola

134k1278176

asked Mar 15 at 17:08

user651754

$endgroup$

put on hold as off-topic by RRL, mrtaurho, Cesareo, José Carlos Santos, Parcly Taxel yesterday

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – RRL, mrtaurho, Cesareo, José Carlos Santos, Parcly Taxel

If this question can be reworded to fit the rules in the help center, please edit the question.

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Maybe you saw it without the bounds like it was $int_0^infty$. Because this way it doesn't look charming.
  $endgroup$
  – Zacky
  Mar 15 at 17:18
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  It's without the bounds in original. I saw it correctly :))))
  $endgroup$
  – user651754
  Mar 15 at 17:22
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @user651754 Can you provide citations for where you saw it?
  $endgroup$
  – tatan
  Mar 15 at 17:22
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  It's a romanian magazine, called ,,Gazeta Matematica". It's from the febraury issue, 2018.
  $endgroup$
  – user651754
  Mar 15 at 17:24
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Zacky It's not a very difficult indefinite integral. See my post.
  $endgroup$
  – Mark Viola
  Mar 15 at 18:47
  

  
  
  
  

add a comment | 

$begingroup$

Find $displaystyleint dfrac{x^nln x}{(x^{n+1}+1)^n}dx$, where $ninmathbb{N}$, $xin(0,infty)$

I think I saw this on mathstack, but I can't find it. Anyway, for this problem, my idea was to consider $x^{n+1}=t$, but that $ln $ put a stop to any valuable development.

calculus integration indefinite-integrals

share|cite|improve this question

edited Mar 15 at 18:48

Mark Viola

134k1278176

asked Mar 15 at 17:08

user651754

$endgroup$

share|cite|improve this question

edited Mar 15 at 18:48

Mark Viola

134k1278176

asked Mar 15 at 17:08

user651754

share|cite|improve this question

edited Mar 15 at 18:48

Mark Viola

134k1278176

asked Mar 15 at 17:08

user651754

edited Mar 15 at 18:48

Mark Viola

134k1278176

edited Mar 15 at 18:48

Mark Viola

134k1278176

1 Answer
1

active

oldest

votes

2

$begingroup$

Let $I(n)$ be given by the integral

$$I(n)=intfrac{x^nlog(x)}{(x^{n+1}+1)^n},dxtag1$$

Integrating by parts the integral in $(1)$ with $u=log(x)$ and $v=-frac{1}{(n-1)(n+1)(x^{n+1}+1)^{n-1}}$ reveals

$$I(n)=-frac{1}{(n+1)(n-1)}frac{log(x)}{(x^{n+1}+1)^{n-1}}+frac{1}{(n+1)(n-1)}int frac1{x(x^{n+1}+1)^{n-1}},dxtag2$$

---

Next, enforcing the substitution $x= y^{1/(n+1)}$ in the integral on the right-hand side of $(2)$, we obtain

$$begin{align}
int frac1{x(x^{n+1}+1)^{n-1}},dx&=frac{1}{n+1}int frac{1}{y(y+1)^{n-1}},dy\\
&=frac1{n+1}int left(frac1y-sum_{k=1}^{n-1} frac1{(y+1)^k}right),dy\
\
&=frac1{n+1}left(log(y)-log(1+y)+sum_{k=2}^{n-1} frac{1}{k-1}frac{1}{(y+1)^{k-1}}right)+C\\
&=log(x)-frac{log(x^{n+1}+1)}{n+1}+frac1{n+1}sum_{k=2}^{n-1} frac{1}{(k-1)(x^{n+1}+1)^k}+C
end{align}$$

---

Can you finish it up now?

share|cite|improve this answer

edited Mar 15 at 18:48

answered Mar 15 at 18:14

Mark ViolaMark Viola

134k1278176

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Indeed, it wasn't that hard. Nice!
  $endgroup$
  – Zacky
  Mar 15 at 18:57
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Zacky Thank you.
  $endgroup$
  – Mark Viola
  Mar 15 at 19:24
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Nice +1. Could I see a proof for $$frac1{y(y+1)^n}=frac1y-sum_{k=0}^{n-1}frac1{(y+1)^k}$$
  $endgroup$
  – clathratus
  Mar 15 at 23:01
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  It is just a partial fraction decomposition formula. Try using the cover up method and see if it works out.
  $endgroup$
  – Abhishek Vangipuram
  Mar 16 at 0:17
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @clathratus You're welcome. My pleasure.
  $endgroup$
  – Mark Viola
  Mar 16 at 0:40
  

  
  
  
  

 | 
show 2 more comments

2

$begingroup$

Let $I(n)$ be given by the integral

$$I(n)=intfrac{x^nlog(x)}{(x^{n+1}+1)^n},dxtag1$$

Integrating by parts the integral in $(1)$ with $u=log(x)$ and $v=-frac{1}{(n-1)(n+1)(x^{n+1}+1)^{n-1}}$ reveals

$$I(n)=-frac{1}{(n+1)(n-1)}frac{log(x)}{(x^{n+1}+1)^{n-1}}+frac{1}{(n+1)(n-1)}int frac1{x(x^{n+1}+1)^{n-1}},dxtag2$$

---

Next, enforcing the substitution $x= y^{1/(n+1)}$ in the integral on the right-hand side of $(2)$, we obtain

$$begin{align}
int frac1{x(x^{n+1}+1)^{n-1}},dx&=frac{1}{n+1}int frac{1}{y(y+1)^{n-1}},dy\\
&=frac1{n+1}int left(frac1y-sum_{k=1}^{n-1} frac1{(y+1)^k}right),dy\
\
&=frac1{n+1}left(log(y)-log(1+y)+sum_{k=2}^{n-1} frac{1}{k-1}frac{1}{(y+1)^{k-1}}right)+C\\
&=log(x)-frac{log(x^{n+1}+1)}{n+1}+frac1{n+1}sum_{k=2}^{n-1} frac{1}{(k-1)(x^{n+1}+1)^k}+C
end{align}$$

---

Can you finish it up now?

share|cite|improve this answer

edited Mar 15 at 18:48

answered Mar 15 at 18:14

Mark ViolaMark Viola

134k1278176

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Indeed, it wasn't that hard. Nice!
  $endgroup$
  – Zacky
  Mar 15 at 18:57
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Zacky Thank you.
  $endgroup$
  – Mark Viola
  Mar 15 at 19:24
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Nice +1. Could I see a proof for $$frac1{y(y+1)^n}=frac1y-sum_{k=0}^{n-1}frac1{(y+1)^k}$$
  $endgroup$
  – clathratus
  Mar 15 at 23:01
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  It is just a partial fraction decomposition formula. Try using the cover up method and see if it works out.
  $endgroup$
  – Abhishek Vangipuram
  Mar 16 at 0:17
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @clathratus You're welcome. My pleasure.
  $endgroup$
  – Mark Viola
  Mar 16 at 0:40
  

  
  
  
  

 | 
show 2 more comments

2

$begingroup$

Let $I(n)$ be given by the integral

$$I(n)=intfrac{x^nlog(x)}{(x^{n+1}+1)^n},dxtag1$$

Integrating by parts the integral in $(1)$ with $u=log(x)$ and $v=-frac{1}{(n-1)(n+1)(x^{n+1}+1)^{n-1}}$ reveals

$$I(n)=-frac{1}{(n+1)(n-1)}frac{log(x)}{(x^{n+1}+1)^{n-1}}+frac{1}{(n+1)(n-1)}int frac1{x(x^{n+1}+1)^{n-1}},dxtag2$$

---

Next, enforcing the substitution $x= y^{1/(n+1)}$ in the integral on the right-hand side of $(2)$, we obtain

$$begin{align}
int frac1{x(x^{n+1}+1)^{n-1}},dx&=frac{1}{n+1}int frac{1}{y(y+1)^{n-1}},dy\\
&=frac1{n+1}int left(frac1y-sum_{k=1}^{n-1} frac1{(y+1)^k}right),dy\
\
&=frac1{n+1}left(log(y)-log(1+y)+sum_{k=2}^{n-1} frac{1}{k-1}frac{1}{(y+1)^{k-1}}right)+C\\
&=log(x)-frac{log(x^{n+1}+1)}{n+1}+frac1{n+1}sum_{k=2}^{n-1} frac{1}{(k-1)(x^{n+1}+1)^k}+C
end{align}$$

---

Can you finish it up now?

share|cite|improve this answer

edited Mar 15 at 18:48

answered Mar 15 at 18:14

Mark ViolaMark Viola

134k1278176

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Indeed, it wasn't that hard. Nice!
  $endgroup$
  – Zacky
  Mar 15 at 18:57
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Zacky Thank you.
  $endgroup$
  – Mark Viola
  Mar 15 at 19:24
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Nice +1. Could I see a proof for $$frac1{y(y+1)^n}=frac1y-sum_{k=0}^{n-1}frac1{(y+1)^k}$$
  $endgroup$
  – clathratus
  Mar 15 at 23:01
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  It is just a partial fraction decomposition formula. Try using the cover up method and see if it works out.
  $endgroup$
  – Abhishek Vangipuram
  Mar 16 at 0:17
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @clathratus You're welcome. My pleasure.
  $endgroup$
  – Mark Viola
  Mar 16 at 0:40
  

  
  
  
  

 | 
show 2 more comments

$begingroup$

Let $I(n)$ be given by the integral

$$I(n)=intfrac{x^nlog(x)}{(x^{n+1}+1)^n},dxtag1$$

Integrating by parts the integral in $(1)$ with $u=log(x)$ and $v=-frac{1}{(n-1)(n+1)(x^{n+1}+1)^{n-1}}$ reveals

$$I(n)=-frac{1}{(n+1)(n-1)}frac{log(x)}{(x^{n+1}+1)^{n-1}}+frac{1}{(n+1)(n-1)}int frac1{x(x^{n+1}+1)^{n-1}},dxtag2$$

---

Next, enforcing the substitution $x= y^{1/(n+1)}$ in the integral on the right-hand side of $(2)$, we obtain

$$begin{align}
int frac1{x(x^{n+1}+1)^{n-1}},dx&=frac{1}{n+1}int frac{1}{y(y+1)^{n-1}},dy\\
&=frac1{n+1}int left(frac1y-sum_{k=1}^{n-1} frac1{(y+1)^k}right),dy\
\
&=frac1{n+1}left(log(y)-log(1+y)+sum_{k=2}^{n-1} frac{1}{k-1}frac{1}{(y+1)^{k-1}}right)+C\\
&=log(x)-frac{log(x^{n+1}+1)}{n+1}+frac1{n+1}sum_{k=2}^{n-1} frac{1}{(k-1)(x^{n+1}+1)^k}+C
end{align}$$

---

Can you finish it up now?

share|cite|improve this answer

edited Mar 15 at 18:48

answered Mar 15 at 18:14

Mark ViolaMark Viola

134k1278176

$endgroup$

share|cite|improve this answer

edited Mar 15 at 18:48

answered Mar 15 at 18:14

Mark ViolaMark Viola

134k1278176

answered Mar 15 at 18:14

Mark ViolaMark Viola

134k1278176

answered Mar 15 at 18:14

Mark ViolaMark Viola

134k1278176