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chrisplusian
 4 years ago
Quick integration question
chrisplusian
 4 years ago
Quick integration question

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chrisplusian
 4 years ago
Best ResponseYou've already chosen the best response.0\[\int\limits x1+\frac{ 1 }{ x+2 } +\frac{ 1 }{ x1 } dx\]

chrisplusian
 4 years ago
Best ResponseYou've already chosen the best response.0is this the same as:

chrisplusian
 4 years ago
Best ResponseYou've already chosen the best response.0\[\int\limits x1 dx +\int\limits \frac{ 1 }{ x+2 } dx +\int\limits \frac{ 1 }{ x1 } dx\] ????

AccessDenied
 4 years ago
Best ResponseYou've already chosen the best response.1Yes, that would be correct.

chrisplusian
 4 years ago
Best ResponseYou've already chosen the best response.0I am doing integration by partial fraction decomposition, I got this far on my own but when I checked wolfram alpha it showed the "x1" as the integral of x dx, and then the integral of 1 dx and somehow their answer and my own do not match

AccessDenied
 4 years ago
Best ResponseYou've already chosen the best response.1You may also split (x  1) into separate integrals, that should work out as well.

chrisplusian
 4 years ago
Best ResponseYou've already chosen the best response.0are these answers the same:

chrisplusian
 4 years ago
Best ResponseYou've already chosen the best response.0\[x^2 2x +1 +2\ln \left x+2 \right +2\ln \left x1 \right\]

chrisplusian
 4 years ago
Best ResponseYou've already chosen the best response.0\[\frac{ 1 }{ 2 }(x2)x +\ln (x1) +\ln (x+2)\]

chrisplusian
 4 years ago
Best ResponseYou've already chosen the best response.0The first one is my answer the second is wolfram alpha

AccessDenied
 4 years ago
Best ResponseYou've already chosen the best response.1Your answer appears to be about the same as wolfram's, except it was multiplied by 2, which doesn't work. The absolute values are fine to have here. Also, since you are doing an indefinite integration, both answers would have an additional arbitrary constant.

chrisplusian
 4 years ago
Best ResponseYou've already chosen the best response.0when I had the integral of x1 dx I substituted and said u=x1 and du=dx. then integrating I ended up with (u^2)/2. plugging back in for u I got the polynomial over two. to clear the fraction from the entire thing I multiplied each term by 2 hence the 2ln.... is that wrong?

AccessDenied
 4 years ago
Best ResponseYou've already chosen the best response.1yeah, that wouldn't work out because you essentially changed the value of the original integral. Multiplying by 2 also multiplies your original integral all by 2.

chrisplusian
 4 years ago
Best ResponseYou've already chosen the best response.0I can see how my answer is similar to wolfram's but, when you distribute the x into the (x2) of wolframs solution you get (x^2)2x an that is all divided by two. My solution was (X^2)2x +1.

chrisplusian
 4 years ago
Best ResponseYou've already chosen the best response.0so you are saying that I can't take\[\frac{ x^2 2x +1 }{ 2 } +\ln \left x+2 \right +\ln \left x1 \right +c\]

chrisplusian
 4 years ago
Best ResponseYou've already chosen the best response.0and multiply through by two to clear the fraction?

AccessDenied
 4 years ago
Best ResponseYou've already chosen the best response.1Yes, since you basically have: \[ \int (x  1 + \frac{1}{x+2} + \frac{1}{x2}) \; dx = \frac{1}{2}(x^2  2x + 1) + \lnx + 2 + \lnx1 + c \] You'd have to multiply both sides by 2 for the equality to hold.

AccessDenied
 4 years ago
Best ResponseYou've already chosen the best response.1So, you'd have a slightly different integral. However, note that your answers vary only by a constant when we don't multiply by 2 (if we add in absolute values to their answer): 1/2(x^2  2x) + lnx+2 + lnx1 + c1 = 1/2(x^2  2x + 1) + lnx+2 + lnx1 + c2 1/2(x^2  2x) + lnx+2 + lnx1 + c1 = 1/2(x^2  2x) + lnx+2 + lnx1 + 1/2 + c2 c1 = 1/2 + c2 < They only vary by constant values Since c2 is just some random constant, we could also just combine 1/2 + c2 into one constant.

chrisplusian
 4 years ago
Best ResponseYou've already chosen the best response.0ok that is kind of confusing. so my answer was right just without the minus two. Is there a way I can use wolfram to see if my answer is equivalent to the one they give me?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0= x²/2  x + ln  x+2 + ln  x 1 + C

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0or x²  2x + 2ln  x+2 + 2ln  x 1 + C

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0You have 4 separates terms to take derivative. The result looks neat without fraction !

chrisplusian
 4 years ago
Best ResponseYou've already chosen the best response.0I am still somewhat confused I don't get how they are the same

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0The first one has fraction 1/2 If you don't want fraction form, you need to multiply all terms by 2!

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0= x²/2  x + ln  x+2 + ln  x 1 + C x²  2x + 2ln  x+2 + 2ln  x 1 + C

AccessDenied
 4 years ago
Best ResponseYou've already chosen the best response.1How would you multiply by 2 without changing the value of the original integral? If you take the derivative of the indefinite integral multiplied by 2, you get the original function, multiplied by 2...

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0This is how you multiply: = 2 ( x²/2  x + ln  x+2 + ln  x 1 ) + C = x²  2x + 2ln  x+2 + 2ln  x 1 + C

chrisplusian
 4 years ago
Best ResponseYou've already chosen the best response.0ok to start with I did that but then @AccessDenied said I could not. The basic problem I have is understanding how wolfram alpha's answer is the same as mine. @Chlorophyll could you scroll up to where i posted my answer as compared to my answer?

chrisplusian
 4 years ago
Best ResponseYou've already chosen the best response.0I got (x^2)2x+1 and I just cannot see how that is the same as (x2)x

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0This is just plain straight forward formula from the textbook!

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0" I got (x^2)2x+1 " Where do you get 1?

chrisplusian
 4 years ago
Best ResponseYou've already chosen the best response.0sorry I must have missed the "plain straightforward" part of my calculus II textbook

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Does it mean you understand your mistake?

AccessDenied
 4 years ago
Best ResponseYou've already chosen the best response.1It is an indefinite integral, so your answer has one constant of variation added on. The difference between your answers is a constant, so it can be combined with your constant of variation since C + 1/2 is also a constant. :P

chrisplusian
 4 years ago
Best ResponseYou've already chosen the best response.0I used the "x1" as u then du=dx. so I then have the integral of u du. That evaluated becomes (u^2)/2. plug x1 back in for u and then square it. that gives you (x^2)2x+1 all divided by two

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0u = x 1 > du = dx => ∫ du/ u = ln u = ln  x1

chrisplusian
 4 years ago
Best ResponseYou've already chosen the best response.0it was just x1 not 1/(x1)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Oh, so they're 2 individual terms!

chrisplusian
 4 years ago
Best ResponseYou've already chosen the best response.0can't you integrate them as I just described? I just ask cause that is where my mind went when I had to do it without any help

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0No, it doesn't work that way. We only apply substitution when we're unable to integral term by term: Suppose ∫ ( x  1) dx Let u = x 1 => du = dx =>∫ udu = u²/2 = ( x1)² /2 = ( x²  2x + 1 ) /2 = x²/2  x + 1/2

AccessDenied
 4 years ago
Best ResponseYou've already chosen the best response.1= x²/2  x + 1/2 [+ c] Let 1/2 + c = C (combine constants into a single constant of variation) = x²/2  x + C

AccessDenied
 4 years ago
Best ResponseYou've already chosen the best response.1The big thing about indefinite integration is that there is an additional constant of variation due to differentiation removing any constants and you are reversing the process. These answers only vary by constants, where differentiation removes both 1/2 and your constant of variation...
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