Evaluate $int x^x^n+n-1(x ln x +1)mathrm dx$
Clash Royale CLAN TAG#URR8PPP
up vote
3
down vote
favorite
$$int x^x^n+n-1(x ln x +1) mathrm dx$$
I tried integration by part, I had no result. can someone help?
calculus integration indefinite-integrals
edited Sep 6 at 8:47
Lorenzo B.
1,5402418
asked Sep 6 at 8:02
user123
546
add a comment |Â
up vote
3
down vote
favorite
$$int x^x^n+n-1(x ln x +1) mathrm dx$$
I tried integration by part, I had no result. can someone help?
calculus integration indefinite-integrals
edited Sep 6 at 8:47
Lorenzo B.
1,5402418
asked Sep 6 at 8:02
user123
546
add a comment |Â
up vote
3
down vote
favorite
up vote
3
down vote
favorite
$$int x^x^n+n-1(x ln x +1) mathrm dx$$
I tried integration by part, I had no result. can someone help?
calculus integration indefinite-integrals
edited Sep 6 at 8:47
Lorenzo B.
1,5402418
asked Sep 6 at 8:02
user123
546
$$int x^x^n+n-1(x ln x +1) mathrm dx$$
I tried integration by part, I had no result. can someone help?
calculus integration indefinite-integrals
calculus integration indefinite-integrals
edited Sep 6 at 8:47
Lorenzo B.
1,5402418
asked Sep 6 at 8:02
user123
546
edited Sep 6 at 8:47
Lorenzo B.
1,5402418
asked Sep 6 at 8:02
user123
546
edited Sep 6 at 8:47
Lorenzo B.
1,5402418
edited Sep 6 at 8:47
Lorenzo B.
1,5402418
edited Sep 6 at 8:47
Lorenzo B.
1,5402418
1,5402418
asked Sep 6 at 8:02
user123
546
asked Sep 6 at 8:02
user123
546
asked Sep 6 at 8:02
user123
546
546
add a comment |Â
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
2
down vote
accepted
This is general form of integral:
$$int x^x(ln x+1)dx= x^x$$
Where n=1. By taking derivative for example for n=2 we can find how to manipulate the integrand to find the integral;
$$y'=x^x^n+n-1(nln x+1)=x^x^n.x^n-1(nln x+1)$$
⇒ $$x^n-1(nln x+1)=n x^n-1 ln x +fracx^nx=(x^n ln x)'=(ln x^x^n)'$$
⇒ $$y'=x^x^n+n-1(nln x+1)=x^x^n.(ln x^x^n)'=$$
Now we compare this with:
$$ y=f(x)⇒ln y =ln f(x)⇒fracy'y=[ln f(x)]'$$
and conclude that:
$$ y=x^x^n$$
answered Sep 6 at 13:23
sirous
1,050512
add a comment |Â
up vote
6
down vote
I think the problem is $$int x^x^n+n-1(n ln x +1) dx$$
If $y=x^x^n$
$ln y=x^nln x$
$$frac1ydfracdydx=nx^n-1ln x+x^n-1$$
$$impliesdfracdydx=x^x^n+n-1(nln x+1)$$
edited Sep 6 at 8:18
GoodDeeds
10.2k21335
answered Sep 6 at 8:12
lab bhattacharjee
216k14153266
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
This is general form of integral:
$$int x^x(ln x+1)dx= x^x$$
Where n=1. By taking derivative for example for n=2 we can find how to manipulate the integrand to find the integral;
$$y'=x^x^n+n-1(nln x+1)=x^x^n.x^n-1(nln x+1)$$
⇒ $$x^n-1(nln x+1)=n x^n-1 ln x +fracx^nx=(x^n ln x)'=(ln x^x^n)'$$
⇒ $$y'=x^x^n+n-1(nln x+1)=x^x^n.(ln x^x^n)'=$$
Now we compare this with:
$$ y=f(x)⇒ln y =ln f(x)⇒fracy'y=[ln f(x)]'$$
and conclude that:
$$ y=x^x^n$$
answered Sep 6 at 13:23
sirous
1,050512
add a comment |Â
up vote
2
down vote
accepted
This is general form of integral:
$$int x^x(ln x+1)dx= x^x$$
Where n=1. By taking derivative for example for n=2 we can find how to manipulate the integrand to find the integral;
$$y'=x^x^n+n-1(nln x+1)=x^x^n.x^n-1(nln x+1)$$
⇒ $$x^n-1(nln x+1)=n x^n-1 ln x +fracx^nx=(x^n ln x)'=(ln x^x^n)'$$
⇒ $$y'=x^x^n+n-1(nln x+1)=x^x^n.(ln x^x^n)'=$$
Now we compare this with:
$$ y=f(x)⇒ln y =ln f(x)⇒fracy'y=[ln f(x)]'$$
and conclude that:
$$ y=x^x^n$$
answered Sep 6 at 13:23
sirous
1,050512
add a comment |Â
up vote
2
down vote
accepted
up vote
2
down vote
accepted
This is general form of integral:
$$int x^x(ln x+1)dx= x^x$$
Where n=1. By taking derivative for example for n=2 we can find how to manipulate the integrand to find the integral;
$$y'=x^x^n+n-1(nln x+1)=x^x^n.x^n-1(nln x+1)$$
⇒ $$x^n-1(nln x+1)=n x^n-1 ln x +fracx^nx=(x^n ln x)'=(ln x^x^n)'$$
⇒ $$y'=x^x^n+n-1(nln x+1)=x^x^n.(ln x^x^n)'=$$
Now we compare this with:
$$ y=f(x)⇒ln y =ln f(x)⇒fracy'y=[ln f(x)]'$$
and conclude that:
$$ y=x^x^n$$
answered Sep 6 at 13:23
sirous
1,050512
This is general form of integral:
$$int x^x(ln x+1)dx= x^x$$
Where n=1. By taking derivative for example for n=2 we can find how to manipulate the integrand to find the integral;
$$y'=x^x^n+n-1(nln x+1)=x^x^n.x^n-1(nln x+1)$$
⇒ $$x^n-1(nln x+1)=n x^n-1 ln x +fracx^nx=(x^n ln x)'=(ln x^x^n)'$$
⇒ $$y'=x^x^n+n-1(nln x+1)=x^x^n.(ln x^x^n)'=$$
Now we compare this with:
$$ y=f(x)⇒ln y =ln f(x)⇒fracy'y=[ln f(x)]'$$
and conclude that:
$$ y=x^x^n$$
answered Sep 6 at 13:23
sirous
1,050512
edited Sep 6 at 15:17
answered Sep 6 at 13:23
sirous
1,050512
answered Sep 6 at 13:23
sirous
1,050512
answered Sep 6 at 13:23
sirous
1,050512
1,050512
add a comment |Â
add a comment |Â
up vote
6
down vote
I think the problem is $$int x^x^n+n-1(n ln x +1) dx$$
If $y=x^x^n$
$ln y=x^nln x$
$$frac1ydfracdydx=nx^n-1ln x+x^n-1$$
$$impliesdfracdydx=x^x^n+n-1(nln x+1)$$
edited Sep 6 at 8:18
GoodDeeds
10.2k21335
answered Sep 6 at 8:12
lab bhattacharjee
216k14153266
add a comment |Â
up vote
6
down vote
I think the problem is $$int x^x^n+n-1(n ln x +1) dx$$
If $y=x^x^n$
$ln y=x^nln x$
$$frac1ydfracdydx=nx^n-1ln x+x^n-1$$
$$impliesdfracdydx=x^x^n+n-1(nln x+1)$$
edited Sep 6 at 8:18
GoodDeeds
10.2k21335
answered Sep 6 at 8:12
lab bhattacharjee
216k14153266
add a comment |Â
up vote
6
down vote
up vote
6
down vote
I think the problem is $$int x^x^n+n-1(n ln x +1) dx$$
If $y=x^x^n$
$ln y=x^nln x$
$$frac1ydfracdydx=nx^n-1ln x+x^n-1$$
$$impliesdfracdydx=x^x^n+n-1(nln x+1)$$
edited Sep 6 at 8:18
GoodDeeds
10.2k21335
answered Sep 6 at 8:12
lab bhattacharjee
216k14153266
I think the problem is $$int x^x^n+n-1(n ln x +1) dx$$
If $y=x^x^n$
$ln y=x^nln x$
$$frac1ydfracdydx=nx^n-1ln x+x^n-1$$
$$impliesdfracdydx=x^x^n+n-1(nln x+1)$$
edited Sep 6 at 8:18
GoodDeeds
10.2k21335
answered Sep 6 at 8:12
lab bhattacharjee
216k14153266
edited Sep 6 at 8:18
GoodDeeds
10.2k21335
edited Sep 6 at 8:18
GoodDeeds
10.2k21335
edited Sep 6 at 8:18
GoodDeeds
10.2k21335
10.2k21335
answered Sep 6 at 8:12
lab bhattacharjee
216k14153266
answered Sep 6 at 8:12
lab bhattacharjee
216k14153266
answered Sep 6 at 8:12
lab bhattacharjee
216k14153266
216k14153266
add a comment |Â
add a comment |Â
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2907212%2fevaluate-int-xxnn-1x-ln-x-1-mathrm-dx%23new-answer', 'question_page');
);
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
