Integrating ln(x+1) with Integration by Parts

[noobie!]

Homework Statement

integrate ln(x+1) dx (integrate by part)

Homework Equations

integrate e^-2x sin3x dx

The Attempt at a Solution

1st,i make u=x+1 ,so du/dx =1 du=dx..while i use /dv =/ln(x+1) dx and after integrating my v=1/(x+1) so subtitute into uv -/vdu but my ans turn out to be wrong so i need some guidance over here.i guess my integration for dv is wrong but i can't really see it..so could any1 please help me..thanks..

 

[NoMoreExams]

How about letting u = ln(x+1) and dv = 1 dx?

I'm not sure what you meant by letting u = x + 1 and dv = ln(x+1). You have integral of ln(x+1) dx, you are subst. u and dv in there, once you let u = x + 1 it makes no sense to let dv = ln(x+1)

 

[rootX]

Homework Statement

integrate ln(x+1) dx (integrate by part)

Homework Equations

integrate e^-2x sin3x dx

The Attempt at a Solution

1st,i make u=x+1 ,so du/dx =1 du=dx..while i use /dv =/ln(x+1) dx and after integrating my v=1/(x+1) so subtitute into uv -/vdu but my ans turn out to be wrong so i need some guidance over here.i guess my integration for dv is wrong but i can't really see it..so could any1 please help me..thanks..

uv -/vdu

/dv =/ln(x+1) dx

if / is integral sign.. then that ln(x+1) is wrong.

It's like

d/dx ( x*ln(x+1)) = ?

and uv = x*ln(x+2)

You just rearrange that "d/dx ( x*ln(x+1)) = ?" equation .. to find int [ln(x+1)] .. I never learned using u and v

Edit: I think NoMoreExams is right... I never used u-v formula

 

[noobie!]

How about letting u = ln(x+1) and dv = 1 dx?

I'm not sure what you meant by letting u = x + 1 and dv = ln(x+1). You have integral of ln(x+1) dx, you are subst. u and dv in there, once you let u = x + 1 it makes no sense to let dv = ln(x+1)

i did like what you mention but i couldn't find the ans..the ans is (x+1)(ln(x+1)) -(x+1) + c ,if i differetiate du/dx =1/(x+1) and sub into the formula i can't get..

 

[noobie!]

uv -/vdu

/dv =/ln(x+1) dx

if / is integral sign.. then that ln(x+1) is wrong.

It's like

d/dx ( x*ln(x+1)) = ?

and uv = x*ln(x+2)

You just rearrange that "d/dx ( x*ln(x+1)) = ?" equation .. to find int [ln(x+1)] .. I never learned using u and v

Edit: I think NoMoreExams is right... I never used u-v formula

ya that's integral sign,why do you say it's wrong?can you please tell me..?thanks..i did like what nomoreexams said..i still can't get my ans correct..i sow you my steps please rectify me.. like what he says; u=ln(x+1) after diff du/dx = 1/x+1 and u integrate dv=1 dx..it will be v= x so finally sub into the formula integration by parts will be uv-/vdu =(ln(x+1))(x)-/(x)(1/x+1 dx) but the ans turn out to be wrong..

 

[rootX]

i did like what you mention but i couldn't find the ans..the ans is (x+1)(ln(x+1)) -(x+1) + c ,if i differetiate du/dx =1/(x+1) and sub into the formula i can't get..

uv-/vdu =(ln(x+1))(x)-/(x)(1/x+1 dx) is NOT equal to (x+1)(ln(x+1)) -(x+1) + c

integral of x/(x+1) is wrong. Are you using partial fractions here?

Edit Sorry,:
>> int (log(x+1),x)

ans =

log(x+1)*(x+1)-x-1

I guess I was making some mistake in the partial fractions.. but you answer looks correct (that's matlab). And, also you can include 1 in c .. it really doesn't make any difference.

 

[noobie!]

uv-/vdu =(ln(x+1))(x)-/(x)(1/x+1 dx) is NOT equal to (x+1)(ln(x+1)) -(x+1) + c

integral of x/(x+1) is wrong. Are you using partial fractions here?

ya,the answer is not match then which part of mine turn out to be wrong?then integrate of 1/(x+1) wrong?why?so what should i do?no,que required to integrate by parts,which means
/u dv = uv - /v du

 

[NoMoreExams]

You guys should look into learning how to typeset in TeX. In any case I'll start you off

\int{ln(x+1) dx} is what we have.

If we let u = ln(x+1) \Rightarrow du = \frac{dx}{x+1} and dv = dx \Rightarrow v = x

Subst. this into our formula we get

\int{ln(x+1) dx = x ln(x+1) - \int \frac{x}{x+1} dx

Can you take it from here?

 

[noobie!]

You guys should look into learning how to typeset in TeX. In any case I'll start you off

\int{ln(x+1) dx} is what we have.

If we let u = ln(x+1) \Rightarrow du = \frac{dx}{x+1} and dv = dx \Rightarrow v = x

Subst. this into our formula we get

\int{ln(x+1) dx = x ln(x+1) - \int \frac{x}{x+1} dx

Can you take it from here?

ok,thanks a lot.,i get your idea..but i can't really take it from there but then i can get it by putting your idea into my way..anyway thank for your help

 

[NoMoreExams]

You can't integrate x/(x+1) dx?

 

[rootX]

You can't integrate x/(x+1) dx?

Yes, I think he had "(x+1)(ln(x+1)) -(x+1) + c " which is right

diff {(x+1)(ln(x+1)) -(x+1) + c }
ln(x+1) + 1 -1 = ln(x+1)

so which is correct. I don't see any problem here other than 1 which can be included in the constant ...

 

[NoMoreExams]

I didn't disagree. I was commenting on him saying "but i can't really take it from there" which meant he couldn't integrate x/(x+1) dx.

 

[noobie!]

I didn't disagree. I was commenting on him saying "but i can't really take it from there" which meant he couldn't integrate x/(x+1) dx.

i could integrate it thanks a lot,yesterday was brain dead..im so sorry,finally i get your points clear d..thanks man,integrate /x/(x+1) will be ln(x+1) i completely understand..thanks to both of you..but i put it in this way,can you please check out could this sol works? 1st,i use a=x+1 ,then integrate lna da i make u=ln a while dv=da so after du/dx=1/a while v=a..from here i depart to sub into the formula uv-/vdu so it will be ...(lna)(a) - /(a) X (1/a da)
then it finally turn out to be ln(x+1)(x+1) - (x+1) + c... does this way work?please do rectify me if there is any mistakes i have made...

 

[NoMoreExams]

i could integrate it thanks a lot,yesterday was brain dead..im so sorry,finally i get your points clear d..thanks man,integrate /x/(x+1) will be ln(x+1) i completely understand..thanks to both of you..but i put it in this way,can you please check out could this sol works? 1st,i use a=x+1 ,then integrate lna da i make u=ln a while dv=da so after du/dx=1/a while v=a..from here i depart to sub into the formula uv-/vdu so it will be ...(lna)(a) - /(a) X (1/a da)
then it finally turn out to be ln(x+1)(x+1) - (x+1) + c... does this way work?please do rectify me if there is any mistakes i have made...

No... integral of x/(x+1) dx is not ln(x+1), that would be integral 1/(x+1) dx.

If you wanted to work with integral of ln(a) da instead of ln(x+1) dx, you can convert it as you said, I don't see why you would but you can.

 

[noobie!]

No... integral of x/(x+1) dx is not ln(x+1), that would be integral 1/(x+1) dx.

If you wanted to work with integral of ln(a) da instead of ln(x+1) dx, you can convert it as you said, I don't see why you would but you can.

oh..ok..im still very weak in integration by parts..do you think of anything which could improve me on that topic?

 

[NoMoreExams]

Practice :) Once you do hundreds of these problems, you'll be able to catch on faster and at least do a few steps in your head. There's also something called tabular integration which you can look up, it simplifies a certain type of integration by parts problems.

 

[noobie!]

Practice :) Once you do hundreds of these problems, you'll be able to catch on faster and at least do a few steps in your head. There's also something called tabular integration which you can look up, it simplifies a certain type of integration by parts problems.

oh,ya sure i tried a lot of these kind of problems..and catching up well,do you have any link which leads to more ques with solutions?by the way thanks for your help,if not i won't be able to get these things solved..haha..

 

[rootX]

oh,ya sure i tried a lot of these kind of problems..and catching up well,do you have any link which leads to more ques with solutions?by the way thanks for your help,if not i won't be able to get these things solved..haha..

which book you are using?
Most calculus books have like +200 integration questions ... You can always go to the library and find Stewart etc.
http://www.stewartcalculus.com/

I remember looking for some practice sites but wasn't really successful.

And, you don't need to know the answers for all the questions you solve.

 

[noobie!]

which book you are using?
Most calculus books have like +200 integration questions ... You can always go to the library and find Stewart etc.
http://www.stewartcalculus.com/

ok,thank you...where are you from?i'm using calculus university as my reference..

 

[rootX]

ok,thank you...where are you from?i'm using calculus university as my reference..

I think Stewart is popular in US/Canada. I was using Glyn James book (a British), but I just went to my library and borrowed some 6 calculus books and then worked on like 2 books (one was Stewart).

 

[noobie!]

I think Stewart is popular in US/Canada. I was using Glyn James book (a British), but I just went to my library and borrowed some 6 calculus books and then worked on like 2 books (one was Stewart).

oh..so you are from Canada? haha..oh well,i will find out those books as you mentioned..which field are you in?engineering?

 

[rootX]

oh..so you are from Canada? haha..oh well,i will find out those books as you mentioned..which field are you in?engineering?

Yes, and .. electrical. I have been using Glyn James for all my calculus classes (I,II,III). It has really good and challenging engineering questions. Mechanical/others people used Calculus for Engineers by Trim. I found it pretty good. Stewart is used mainly for non-engineering math courses (at my university) but it can be very useful for first year (engineering) calculus courses.

 

[noobie!]

Yes, and .. electrical. I have been using Glyn James for all my calculus classes (I,II,III). It has really good and challenging engineering questions. Mechanical/others people used Calculus for Engineers by Trim. I found it pretty good. Stewart is used mainly for non-engineering math courses (at my university) but it can be very useful for first year (engineering) calculus courses.

oh i see..i'm from foundation in engineering,next year going for electrical most probably..haha, that will be very good if you can be one of my mentor..haha,do you have a messenger account?i will try out those questions since its so challenging..thanks a lot anyway..

 

[NoMoreExams]

The best advice I can offer you is to get one of those Calculus for Dummies or Cliffs Notes Calculus or even an AP Calculus (probably BC) study guide. You can check out REA's Problem Solvers Calculus but as I recall it wasn't that great. You can also try googling old calculus tests but that requires quite a bit of research. Stewart is a good suggestion.

 

[noobie!]

The best advice I can offer you is to get one of those Calculus for Dummies or Cliffs Notes Calculus or even an AP Calculus (probably BC) study guide. You can check out REA's Problem Solvers Calculus but as I recall it wasn't that great. You can also try googling old calculus tests but that requires quite a bit of research. Stewart is a good suggestion.

ok..no problem..thanks for guiding me all the way..

 

[rootX]

haha,do you have a messenger account?i will try out those questions since its so challenging..thanks a lot anyway..

hmm... life's too busy so I can't afford it. Sorry

 

[Gregg]

Homework Statement

integrate ln(x+1) dx (integrate by part)

Homework Equations

integrate e^-2x sin3x dx

The Attempt at a Solution

1st,i make u=x+1 ,so du/dx =1 du=dx..while i use /dv =/ln(x+1) dx and after integrating my v=1/(x+1) so subtitute into uv -/vdu but my ans turn out to be wrong so i need some guidance over here.i guess my integration for dv is wrong but i can't really see it..so could any1 please help me..thanks..

\int e^{-2x}sine(3x) \delta x
\int e^{-2x}sine(3x) \delta x = uv - \int v\frac{\delta u}{\delta x} \delta x
\int e^{-2x}sine(3x) \delta x = -\frac{1}{3}e^{-2x}cos(3x)-\frac{2}{3}\int e^{-2x}cos(3x)
\int e^{-2x}sine(3x) \delta x = -\frac{1}{3}e^{-2x}cos(3x)-\frac{2}{3}\left[uv - \int v\frac{\delta v}{\delta x} \delta x\right]
\int e^{-2x}sine(3x) \delta x = -\frac{1}{3}e^{-2x}cos(3x)-\frac{2}{3}\left[-\frac{1}{3}e^{-2x}sin(3x) + \frac{2}{3}\int e^{-2x}sin(3x) \delta x\right]

\int e^{-2x}sine(3x) \delta x = -\frac{1}{3}e^{-2x}cos(3x)-\frac{2}{9}e^{-2x}sin(3x)-\frac{4}{9}\int e^{-2x}sin(3x) \delta x
\frac{13}{9}\int e^{-2x}sin(3x) \delta x = -\frac{1}{3}e^{-2x}cos(3x)-\frac{2}{9}e^{-2x}sin(3x)
\int e^{-2x}sin(3x) \delta x = \frac{9}{13}\left[-\frac{1}{3}e^{-2x}cos(3x)-\frac{2}{9}e^{-2x}sin(3x)\right]


\int e^{-2x}sin(3x) \delta x = -\frac{1}{13}e^{-2x}\left[3cos(3x)+2sin(3x)\right]

That's integration by parts for the part of the problem in bold. I'm not sure why it is needed but maybe if you are troubled by integration by parts those steps make it a lot clearer. As for integrating \int ln|x+1| \delta x it's simpler than the other one.

\int ln|x+1| \delta x
\int ln|x+1| \delta x = uv - \int v \frac{\delta u}{\delta x} \delta x
\int ln|x+1| \delta x = xln|x+1| - \int x\frac{1}{x+1}

u = (x+1)

\int ln|x+1| \delta x = xln|x+1| - \int (u-1)u^{-1} \delta u
\int ln|x+1| \delta x = xln|x+1| - \int 1-\frac{1}{x+1} \delta x
\int ln|x+1| \delta x = xln|x+1| - \left[x - ln|x+1| \right]
\int ln|x+1| \delta x = xln|x+1| + ln|x+1| - x
\int ln|x+1| \delta x = (x+1)ln|x+1| - x

I think that's how you do it.

 

[CompuChip]

I see that usually, partial integration is taught with u, du, v and dv's.
Perhaps it also helps to see another way, by which I have always remembered it (of course, it's equivalent, but it may be easier for some).

In this case, partial integration is formulated
\int f' g dx = f g - \int f g' dx
which means
\int_a^b f'(x) g(x) dx = f(b) g(b) - f(a) g(a) - \int_a^b f(x) g'(x) dx.
Basically, all you have to do is find one function f' which is easy to integrate (or at least, you know you should be able to do it ). Then integrate it, multiply by the other one, and fill in the boundary. Write down a minus sign, integral, the integrated function f, differentiate the other one (which is usually the easy part) and close with dx.

Example:
\int_0^1 x dx = ?
by partial integration. Well, let's write the integrand trivially as 1 * x.
You can integrate 1 to x, and differentiate x to 1, then
\int_0^1 1 x dx = x^2|_0^1 - \int_0^1 x 1 dx.
Note how the second integral is the same as the first one, so we can take it to the other side:
2 \int_0^1 1 x dx = x^2|_0^1:
\int_0^1 1 x dx = \frac12 x^2|_0^1,
which of course you already knew.

All it now comes down to is to decide which function you will be integrating, and which one you will be differentiating. As a rule of thumb, always pick the easiest one for integrating, because integrating is hard and it's easier to differentiate something complicated than integrate it :)

 

[noobie!]

ok.thanks alot,i understand much better..

 

[noobie!]

thanks greg ,i got you..

 

[noobie!]

which book you are using?
Most calculus books have like +200 integration questions ... You can always go to the library and find Stewart etc.
http://www.stewartcalculus.com/

I remember looking for some practice sites but wasn't really successful.

And, you don't need to know the answers for all the questions you solve.

hmm..rootx..do we have a special formula to factorize out for cubic equation?for example (x^3 + 27)?

 

[NoMoreExams]

Yes... you should know that formula and if you do not at the calculus level you should know how to figure it out :)

Note that if f(x) = x^3 + 27 then f(-3) = 0 hence -3 is a root of that equation and therefore (x + 3) divides x^3 + 27. Now do division.