How to derive a general formula using integration by parts?
Q. Using integration by parts, derive a general formula for integral((x^n)(e^(kx)))dx , when k does not equal 0, in which the resulting integrand involoves x^(n-1)
Asked by Rachel H - Tue Sep 25 11:18:16 2007 - - 2 Answers - 0 Comments

A. (1/k) (x^n) de^(kx) = (1/k)(x^n) e^(kx) - (n/k) x^(n-1)e^(kx)dx --- To the answerer below me, How did you get e^(-kx)? It's obviously wrong.
Answered by sahsjing - Tue Sep 25 11:24:41 2007

How to use integration by parts to integrate integrals?
Q. I have to use integration by parts to integrate each of the integrals for x ln(x + 1) dx
Asked by NoahR - Tue Oct 27 16:33:36 2009 - - 2 Answers - 0 Comments

A. When you do integration by parts, you can use mental substitution which is much simpler. I presented the skills in MathFest to AP Calculus teachers several years ago in San Jose, CA. I hope you enjoy this kind of approach. x ln(x + 1) dx = (1/2) ln(x + 1) dx^2 = (1/2)x^2 ln(x+1) - (1/2) x^2/(x+1) dx = (1/2)x^2 ln(x+1) - (1/2) (x^2+2x+1 - 2x-1)/(x+1) dx = (1/2)x^2 ln(x+1) - (1/2) [(x+1) - 2 + 1/(x+1)] dx = (1/2)x^2 ln(x+1) - (1/2)[(1/2)(x+1)^2 - 2x + ln(x+1)] = (1/4)[(2x^2-1)ln(x+1) - (x+1)^2 + 4x] + c --- Attn: My answer has been checked with --- To the answerer below me, Your answer " ( x / (x + 1)) + ln(x + 1) + C " is incorrect. Actually, you can take derivative to check it. [( x / (x + 1)) + ln(x + 1) + C ]' = [1 - 1 / ( [cont.]
Answered by sahsjing - Tue Oct 27 16:46:02 2009

How do you evaluate the integral of sin(t) sin(2t) dt using integration by parts?
Q. I know integration by parts deals with u, du, v, and dv, but I don't know what is what and how to solve it using this method.
Asked by jimmyjohnson - Wed Mar 18 23:14:34 2009 - - 1 Answers - 0 Comments

A. This method uses the fact that you can define the derivative of a the product of a pair of functions to be: d(u*v) = u*dv + v*du You then try to set up the expression to be integrated so that it equals u*dv so that: u*dv = d(uv) - v*du Now the first term d(u*v) is easily integrable to u*v. And if the method is good v*du will also be integrable although some times you may have to apply "parts" more than once in order to get an answer. For this particular problem. You have trig functions of t and 2t but you would like functions of just t as this is easier to work with. sin(2t) = 2*sin(t)*cos(t) ... standard trig identity sin(t)[2*sin(t)cos(t)]dt = 2*sin(^2(t)*cos(t)dt And this is directly integrable without resorting to integration… [cont.]
Answered by Captain Mephisto - Wed Mar 18 23:56:03 2009

I need to practice integration by parts?
Q. I need to practice with the technique of integration by parts. Can someone provide me 5 to 10 problems that I can use to practice? They need to be so that they can be solved without a calculator. Thanks so much!
Asked by Sarah G - Sat Feb 7 17:35:08 2009 - - 4 Answers - 0 Comments

A. Hi. Try these: This has solutions: And some more: Hope this helped!
Answered by F - Sat Feb 7 17:43:27 2009

Can someone help me solve this integration problem using integration by parts?
Q. find the indefinite integral of (cos(2x))^-1 using integration by parts. please show steps. your help is much appreciated.
Asked by Brooke M - Thu Apr 2 15:40:24 2009 - - 2 Answers - 0 Comments

A. The formula for integration by-parts is (int) udv=u*v-(int)vdu. Soo. Let u=cos inverse of 2x; du=-2x/(sqrt) (1-4x^2) dv=dx; so v=x. Plug into said formula. You should get another integral that you can do tabular on. If you don't know tabular (where you differentiate one down to 0 and integrate the other one as many times as you had to differentiate the other one down to 0), just do by-parts again. You should get 2x*cosinverse of 2x - .5(sin inv. 2x) Hope that helped.
Answered by Star 23 - Thu Apr 2 15:52:33 2009

when is it a good idea to use integration by parts?
Q. What clues tell you that you should use the "integration by parts" technique to integrate?
Asked by Jake P - Wed Feb 6 18:59:54 2008 - - 1 Answers - 0 Comments

A. Please visit this website: Scroll to the bottom of the page and click on the examples. It gives you the common cases on when to use it.
Answered by MsMath - Thu Feb 7 11:35:51 2008

How do you use integration by parts to find this integral?
Q. INT (e to 1) (lnx) / (x^2) dx I don't get how you use integration by parts here...any help??
Asked by SoxBacktoBack - Mon Aug 24 22:08:23 2009 - - 1 Answers - 0 Comments

A. Let u = ln(x), dv = 1/x . Then du = 1/x and v = 1/x u dv = uv v du ln(x) dx/x = ln(x)/x dx/x = ln(x)/x 1/x + C = (ln(x) + 1)/x + C The integral from e to 1 is then (ln(1) + 1)/1 + (ln(e) + 1)/e = (0 + 1)/1 + (1 + 1)/e = 2/e 1 = (2 e)/e
Answered by unknown - Mon Aug 24 23:28:54 2009

How could I solve following question using integration by parts?
Q. how could i solve following indefinite integral by parts... int (xe^x)/(x+1)^2 dx please solve it...
Asked by Qasim - Tue Oct 13 15:22:14 2009 - - 1 Answers - 0 Comments

A. xe^x / (x + 1)^2 dx = x*e^x * 1/(x + 1)^2 dx Let: u = x*e^x dv = 1/(x + 1)^2 du = x'*e^x + x*e^x' = e^x + x*e^x = (x + 1)e^x dx v = -1/(x + 1) By integration by parts, we get: uv - v du = -x*e^x/(x + 1) - -1/(x + 1) * (x + 1)e^x dx = -x*e^x/(x + 1) - -e^x dx = -x*e^x/(x + 1) + e^x + C = -x*e^x/(x + 1) + [e^x(x + 1)]/(x + 1)] + C = [-x*e^x + e^x(x + 1)]/(x + 1) + C = [(-x + x + 1)e^x]/(x + 1) + C = e^x/(x + 1) + C I hope this helps!
Answered by unknown - Tue Oct 13 15:39:41 2009

How do I use integration by parts to find the integral -2x ln(x) dx?
Q. I always get confused when working with ln.
Asked by elyssettte - Fri Dec 5 10:08:09 2008 - - 2 Answers - 0 Comments

A. Well assume that you don't know that ln(x) dx = xlnx - x + C, so you have to differentiate it. u = lnx du = (1/x) dx dv = -2x dx v = -x -x lnx - -x (1/x) dx -x lnx + x dx -x lnx + (1/2)x +C (1/2)x (1 - 2lnx) + C
Answered by o - Fri Dec 5 21:23:39 2008

How do you determine which is U and which is DV in solving integration by parts?
Q. How do you solve the integral of (e^x - x)^2 dx and the integral of xln2x dx ? How can you use uv - integral of vdu to solve these problems? your help is greatly appreciated
Asked by bellenrose - Sat Aug 30 15:00:13 2008 - - 1 Answers - 0 Comments

A. Try this: This is the order in which you should pick 'u', the the other is 'dv'. Log Inverse trig Algebraic Trig Exponential Meaning that "u" should always be log first, then inverse trig, then algebraic terms, then trig, and last exponential. For the first integral: Let's foil : (e^x - x)(e^x - x) (e^2x - 2xe^x + x ) dx As you can see, we can split this up into 3 different integrals, making it less likely to make mistakes. Then, when we do integrate all three, we can simply add them back together. e^2x dx here, let u = 2x, so that du = 2 dx So all we need is to put a 2 beside the dx, and we'll have du. We must also put a 1/2 out front, so that we don't change the integral. Thus, 1/2 e^u du = 1/2 e^(2x) <~~ This is… [cont.]
Answered by THE nerd. - Sat Aug 30 15:06:08 2008

Use integration by parts to prove the reduction formula?
Q. Use integration by parts to prove the reduction formula: (ln x)^n dx = x(ln x)^n - n (ln x)^n-1 dx clear, precise step by step please Thank you in advanced
Asked by jay - Tue Oct 6 00:42:43 2009 - - 2 Answers - 0 Comments

A. (ln x)^n dx Taking u = (ln x)^n and v = 1 Integral = u v dx - [du/dx vdx] dx = (ln x)^n dx - [d/dx (ln x)^n dx] dx = x (ln x)^n - [n (ln x)^(n - 1) * d/dx (ln x) * x] dx = x (ln x)^n - [n (ln x)^(n - 1) * (1/x) * x] dx = x (ln x)^n - n (ln x)^(n - 1) dx.
Answered by Madhukar Daftary - Tue Oct 6 00:51:46 2009

How do you integrate x^3*(1+x^2)^(1/2) using integration by parts?
Q. It's a 3-part question and I can do it using trig substitution and change of variable substitution, but I'm getting stuck trying to use IBP. Help would be appreciated. Thanks.
Asked by Kyle - Sun Nov 30 23:30:32 2008 - - 1 Answers - 0 Comments

A. x (x + 1) dx = rewrite it as: x [x (x + 1)] dx = divide and multiply by 2 so as to get (2x) that is the derivative of the radicand: (1/2) x [2x (x + 1)] dx = let: x = u 2x dx = du [2x (x + 1)] dx = dv express the root as a fractionary power: [(x + 1)^(1/2)] 2x dx = dv [(x + 1)^(1/2)] d(x + 1) = dv {(x + 1)^[(1/2) +1]} / [(1/2) +1] = v [(x + 1)^(3/2)] /(3/2) = v (2/3) (x + 1) = v thus, integrating by parts, you get: u dv = v u - v du (1/2) x [2x (x + 1)] dx = (1/2) [x (2/3) (x + 1) - (2/3) (x + 1) 2x dx] x (x + 1) dx = (1/2)(2/3)x (x + 1) - (1/2)(2/3) (x + 1) d(x + 1) x (x + 1) dx = (1/3)x (x + 1) - (1/3) [(x + 1)^(3/2)] d(x + 1) … [cont.]
Answered by germano - Mon Dec 1 00:17:05 2008

How do you use integration by parts to evaluate the integral of inverse cot 3x dx?
Q. How do you use integration by parts to evaluate the integral of inverse cot 3x dx?
Asked by Casey C - Sun Jan 25 13:03:35 2009 - - 1 Answers - 0 Comments

A. Hey mate, int (arccot(x) dx) Let x = cot(u) --> dx/du = -cosec^2(u) --> dx = -cosec^2(u)du Thus, int(arccot(x) dx) = int( arccot(cot(u)) * -cosec^2(u) du) = int(u*(-cosec^2(u)) du) now employ integration by parts i.e. int (ab') = ab - int(ba') Here let, a = u --> a' = 1 b' = -cosec^2(u) --> b = cot(u) Thus, int(arccot(x) dx) = int(u*(-cosec^2(u)) du) = u*cot(u) - int(cot(u) du) note int(cot(u)) = ln(sin(u)) Thus, int(arccot(x) dx) = u*cot(u) - ln(sin(u)) + C Recall x = cot(u) --> u = arccot(x) Next step is to resolve sin(u) in terms of x we know, cot(u) = x --> cos(u)/sin(u) = x --> sqrt( 1 - sin^2(u))/sin(u) = x --> (1-sin^2(u))/sin^2(u) = x^2 (squared both sides) --> (1 - sin^2(u)) = x^2 * sin^2(u) --> sin^2(u) (x^2 + 1) = 1 --> sin^ [cont.]
Answered by unknown - Sun Jan 25 13:22:48 2009

What's a good way to know when to simply use: Substitution only vs. Integration by Parts.?
Q. It's hard to explain this, but sometimes attacking certain problems, your not really sure what to do. Sometimes I think I'm supposed to do Integ. by parts but instead all I had to do was subtitute. Any suggestions?
Asked by Mark - Fri Feb 22 08:08:19 2008 - - 2 Answers - 0 Comments

A. I agree with the previous answer. It is often a matter of experience just knowing what is likely to work because you've done so many in the past. The key to integration by parts is looking for a product of two functions. One must get simpler when you differentiate it (like 5x or x^2) and the other must get no harder when you integrate it (like e^x or cosx). There are exceptions of course but it's not a bad thing to keep in mind. On the other hand substitution is usually involved if there are brackets to a power or square roots around. Again there are exceptions. That's why integration is so difficult. Also remember, there are many functions that simply can't be integrated to a finite sum of basic functions.
Answered by mathsmanretired - Fri Feb 22 09:34:29 2008

Another integration by parts question?
Q. Hi again, I have found another integration by parts question which I am finding quite difficult. (Thanks soo much to the people how replied to my last question) this one is integrate: xln(x-2). Any help would be greatly appreciated.
Asked by shhusan - Wed Sep 10 08:43:39 2008 - - 1 Answers - 0 Comments

A. f(x).g(x).dx = f(x) g(x).dx - [f '(x). g(x).dx].dx taking f(x) as ln[x - 2] and g(x) as x g(x).dx = x.dx = (1/2)x^2 f '(x) = d[ln|x - 2|]/dx = 1/(x - 2) x.ln[x-2] = ln[x-2] xdx - [{d(ln[x-2])/dx}. x.dx].dx ln[x-2] xdx - [(x^2/2)(x-2)].dx ln[x-2] xdx - 1/2 [(x^2)(x-2)].dx ln[x-2] xdx - 1/2 [(x^2 - 4 + 4)/(x-2)].dx ln[x-2] xdx - 1/2 (x^2 - 4)dx/(x-2) - 4 [1/(x-2)].dx ln[x-2] xdx - 1/2 (x - 2)(x + 2)dx/(x-2) - 4/2 [1/(x-2)].dx ln[x-2] xdx - 1/2(x + 2)dx - 2 [1/(x-2)].dx ln[x-2](x /2) - x /4 - x - 2ln[x-2] + c
Answered by Bi Rain & Song Hye Gyo ROCK ! - Wed Sep 10 09:07:27 2008

How do you distinguish with method to use (substitution vs integration by parts) when finding the integral?
Q. How do you distinguish with method to use (substitution vs integration by parts) when finding the integral?
Asked by 00 - Wed Nov 11 20:06:40 2009 - - 1 Answers - 0 Comments

How do you integrate tan^-1 x dx using subsitution and integration by parts with x=tanu ?
Q. How do you integrate tan^-1 x dx using subsitution and integration by parts with x=tanu ?
Asked by Scott D - Sat Feb 7 16:48:49 2009 - - 2 Answers - 0 Comments

A. if you make the substitution x=tan(u), then you get dx=sec^2(u)du. your integral then becomes tan^-1(tan(u))*sec^2(u) du, which simplifies to u*sec^2(u) du. remember that integration by parts tells you w dv = wv- v dw. you can say that dv=sec^(u)du, so v=tan(u), and you can say that w=u, so dw=du. edit: the other guy gave a better way of doing this, so only use mine if you insist on making the x=tan(u) substitution
Answered by hanging-with-db-cooper - Sat Feb 7 16:56:48 2009

Calculus 2 question about integration by parts and the reduction formula.?
Q. How do we know what to substitute in for the values of u,du, v, and dv when using the reduction formula. I know the question is vague and too general, but anything wil do. I just need a hint and ill get it. I know dat in some cases we have to just prove the reduction formula, but in any case, I dont know what to substitute in for those for variables so i can use the formula. anything will do.
Asked by gazan8one8 - Mon Jul 14 02:22:32 2008 - - 2 Answers - 0 Comments

A. I'm not sure which reduction formula you mean... But for integration by parts ( u dv = uv - v du) you can use the acronym LIATE to choose what to let u equal (and what's left over is dv :) Log Inverse Algebraic (i.e. powers of x, radicals) Trigonometric Exponential in that order. For instance, for e^x (x^2 + 2x) dx, A comes before E in LIATE so try u = x^2 + 2x and dv = e^x dx
Answered by a +b =c - Mon Jul 14 02:32:56 2008

How do you use integration by parts and trig identities to solve the integral of cos(3x)sind(2x) ????
Q. How do you use integration by parts and trig identities to solve the integral of cos(3x)sind(2x) ???
Asked by azncat25 - Tue Jan 29 10:46:54 2008 - - 2 Answers - 0 Comments

A. If I = INT:( u ) dv = uv - INT: ( v ) du I = cos(3x)sin(2x)dx If: u = sin(2x) , du = 2cos(2x)dx dv = cos(3x)dx , v = (1/3)sin(3x) So I = (1/3)sin(2x)sin(3x) - (2/3)INT:sin(3x)cos(2x)dx I(2) = INT:sin(3x)cos(2x)dx u = cos(2x) , du = -2sin(2x)dx dv = sin(3x)dx , v = -(1/3)cos(3x) I(2) = -(1/3)cos(2x)cos(3x) - (2/3)*I So I = (1/3)sin(2x)sin(3x) -(2/3)[-(1/3)cos(2x)cos(3 x) - (2/3)*I] I = (1/3)sin(2x)sin(3x) + (2/9)cos(2x)cos(3x)+(4/9) *I (5/9)*I = (1/3)sin(2x)sin(3x)+(2/9) cos(2x)cos(3x) I = (3/5)sin(2x)sin(3x) + (2/5)cos(2x)cos(3x) I = (1/5)[ 2cos(2x)cos(3x)] - (3/10)[-2sin(2x)sin(3x)] Trig Identities are 2cosAcosB = cos(A+B) + cos(A-B) and -2sinAsinB = cos(A+B) - cos(A-B) Note cos(-x) = cos(x) so I = (1/5)[cos(5x) + cos(x) - ( [cont.]
Answered by eazylee369 - Tue Jan 29 11:48:11 2008

Using integration by parts how do you integrate the indef integral of e^(-x)cos2x dx ?
Q. can you show me steb by step. I used integration by parts 2 times? do I need to? should u =e^-x and du=-e^-x and dv =cos2x and v=1/2sin2x?
Asked by houstonman20042002 - Mon Oct 8 00:51:40 2007 - - 2 Answers - 0 Comments

A. I = e^(-x)cos2x dx u = e^(-x) du = -e^(-x)dx dv = cos(2x)dx v = (1/2)sin(2x) udv = uv - vdu = (1/2)sin(2x)e^(-x) + (1/2)sin(2x)e^(-x)dx u= e^(-x) du = -e^(-x)dx dv = (1/2)sin(2x) v = -(1/4)cos(2x) I = (1/2)sin(2x)e^(-x) + (1/2)sin(2x)e^(-x)dx = (1/2)sin(2x)e^(-x) - (1/4)cos(2x)e^(-x) - (1/4) e^(-x)cos(2x)dx = (1/4)e^(-x)[2sin(2x) - cos(2x)] - I/4 + k 5I/4 = (1/4)e^(-x)[2sin(2x) - cos(2x)] + k I = (1/5)e^(-x)[2sin(2x) - cos(2x)] + c
Answered by gudspeling - Mon Oct 8 01:09:24 2007