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Find $int fracx^nln x(x^n+1+1)^ndx$,

How find this integral $intfracx^2+2x+1+(3x+1)sqrtx+lnxxsqrtx+lnx(x+sqrtx+lnx)dx$Evaluating: $int 3xsinleft(frac x4right) , dx$.Compute $intfracdxsqrttan x$Integrate $intfracdx(x^2+16)^3$Computing the Integral $int fracsqrtxx^2+x dx$Integrating $intfracx^2-1(x^2+1)sqrtx^4+1,dx$A problem in calculating integralFinding the integral $int sqrt1-frac3x+frac1x^2 , dx$How can we find a closed form solution of $int(x+a)^m(x+b)^n~dx$?Find the value of $int _0 ^ infty f(x+frac1x)fracln xx dx$

1

1

$begingroup$

Find $displaystyleint dfracx^nln x(x^n+1+1)^ndx$, where $ninmathbbN$, $xin(0,infty)$

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

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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
  

  
  
  
  

 | 
show 1 more comment

1

1

$begingroup$

Find $displaystyleint dfracx^nln x(x^n+1+1)^ndx$, where $ninmathbbN$, $xin(0,infty)$

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

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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
  

  
  
  
  

 | 
show 1 more comment

$begingroup$

Find $displaystyleint dfracx^nln x(x^n+1+1)^ndx$, where $ninmathbbN$, $xin(0,infty)$

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

$$I(n)=intfracx^nlog(x)(x^n+1+1)^n,dxtag1$$

Integrating by parts the integral in $(1)$ with $u=log(x)$ and $v=-frac1(n-1)(n+1)(x^n+1+1)^n-1$ reveals

$$I(n)=-frac1(n+1)(n-1)fraclog(x)(x^n+1+1)^n-1+frac1(n+1)(n-1)int frac1x(x^n+1+1)^n-1,dxtag2$$

---

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

$$beginalign
int frac1x(x^n+1+1)^n-1,dx&=frac1n+1int frac1y(y+1)^n-1,dy\\
&=frac1n+1int left(frac1y-sum_k=1^n-1 frac1(y+1)^kright),dy\
\
&=frac1n+1left(log(y)-log(1+y)+sum_k=2^n-1 frac1k-1frac1(y+1)^k-1right)+C\\
&=log(x)-fraclog(x^n+1+1)n+1+frac1n+1sum_k=2^n-1 frac1(k-1)(x^n+1+1)^k+C
endalign$$

---

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 $$frac1y(y+1)^n=frac1y-sum_k=0^n-1frac1(y+1)^k$$
  $endgroup$
  – 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

Your Answer

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2

$begingroup$

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

$$I(n)=intfracx^nlog(x)(x^n+1+1)^n,dxtag1$$

Integrating by parts the integral in $(1)$ with $u=log(x)$ and $v=-frac1(n-1)(n+1)(x^n+1+1)^n-1$ reveals

$$I(n)=-frac1(n+1)(n-1)fraclog(x)(x^n+1+1)^n-1+frac1(n+1)(n-1)int frac1x(x^n+1+1)^n-1,dxtag2$$

---

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

$$beginalign
int frac1x(x^n+1+1)^n-1,dx&=frac1n+1int frac1y(y+1)^n-1,dy\\
&=frac1n+1int left(frac1y-sum_k=1^n-1 frac1(y+1)^kright),dy\
\
&=frac1n+1left(log(y)-log(1+y)+sum_k=2^n-1 frac1k-1frac1(y+1)^k-1right)+C\\
&=log(x)-fraclog(x^n+1+1)n+1+frac1n+1sum_k=2^n-1 frac1(k-1)(x^n+1+1)^k+C
endalign$$

---

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 $$frac1y(y+1)^n=frac1y-sum_k=0^n-1frac1(y+1)^k$$
  $endgroup$
  – 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

$$I(n)=intfracx^nlog(x)(x^n+1+1)^n,dxtag1$$

Integrating by parts the integral in $(1)$ with $u=log(x)$ and $v=-frac1(n-1)(n+1)(x^n+1+1)^n-1$ reveals

$$I(n)=-frac1(n+1)(n-1)fraclog(x)(x^n+1+1)^n-1+frac1(n+1)(n-1)int frac1x(x^n+1+1)^n-1,dxtag2$$

---

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

$$beginalign
int frac1x(x^n+1+1)^n-1,dx&=frac1n+1int frac1y(y+1)^n-1,dy\\
&=frac1n+1int left(frac1y-sum_k=1^n-1 frac1(y+1)^kright),dy\
\
&=frac1n+1left(log(y)-log(1+y)+sum_k=2^n-1 frac1k-1frac1(y+1)^k-1right)+C\\
&=log(x)-fraclog(x^n+1+1)n+1+frac1n+1sum_k=2^n-1 frac1(k-1)(x^n+1+1)^k+C
endalign$$

---

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 $$frac1y(y+1)^n=frac1y-sum_k=0^n-1frac1(y+1)^k$$
  $endgroup$
  – 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

$$I(n)=intfracx^nlog(x)(x^n+1+1)^n,dxtag1$$

Integrating by parts the integral in $(1)$ with $u=log(x)$ and $v=-frac1(n-1)(n+1)(x^n+1+1)^n-1$ reveals

$$I(n)=-frac1(n+1)(n-1)fraclog(x)(x^n+1+1)^n-1+frac1(n+1)(n-1)int frac1x(x^n+1+1)^n-1,dxtag2$$

---

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

$$beginalign
int frac1x(x^n+1+1)^n-1,dx&=frac1n+1int frac1y(y+1)^n-1,dy\\
&=frac1n+1int left(frac1y-sum_k=1^n-1 frac1(y+1)^kright),dy\
\
&=frac1n+1left(log(y)-log(1+y)+sum_k=2^n-1 frac1k-1frac1(y+1)^k-1right)+C\\
&=log(x)-fraclog(x^n+1+1)n+1+frac1n+1sum_k=2^n-1 frac1(k-1)(x^n+1+1)^k+C
endalign$$

---

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