When finding the integration of (ln x)^2 dx
and letting u = (ln x)^2 dv = dx du = 2(ln x) * 1/x v = x
and following integration of u dv = uv - integration of v du
(ln x)^2 dx = x * (ln x)^2 - integration of x * 2(ln x) * 1/x dx
" " = x(ln x)^2 - [2 * integration of ln x dx]
My knowledge is such that I want to say the answer is x(ln x)^2 - 2/x + C
However, the solutions manual (_Calculus: Early Transcendentals_ 5th ed. James Stewart) goes an extra step and integrates by parts *again* (ln x) and dx, so the final answer is x(ln x)^2 -2xln x +2x +C
I would like to know why the extra step? How would I recognize that my solution was incomplete?
and letting u = (ln x)^2 dv = dx du = 2(ln x) * 1/x v = x
and following integration of u dv = uv - integration of v du
(ln x)^2 dx = x * (ln x)^2 - integration of x * 2(ln x) * 1/x dx
" " = x(ln x)^2 - [2 * integration of ln x dx]
My knowledge is such that I want to say the answer is x(ln x)^2 - 2/x + C
However, the solutions manual (_Calculus: Early Transcendentals_ 5th ed. James Stewart) goes an extra step and integrates by parts *again* (ln x) and dx, so the final answer is x(ln x)^2 -2xln x +2x +C
I would like to know why the extra step? How would I recognize that my solution was incomplete?
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Re: Integration into Parts question
Wed, October 3, 2007 - 12:42 PMShort answer: to figure out the integral of ln(x) dx you need to integrate by parts.
In your solution, it looks like your [2*integration of ln x dx] became 2/x ... but 2/x is the derivative of ln(x), not the integral. -
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Re: Integration into Parts question
Mon, October 15, 2007 - 7:26 AMOof. thank you. For some reason my brain thought it was a commutative property...
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Re: Integration into Parts question
Mon, March 31, 2008 - 5:37 PMas long as there is still a product of functions in the integrand that can't be expressed as udu, we still need to apply integration by parts.
Integration by parts is just the integration equivalent of the product rule for derivatives. Take the product rule statement that:
(u v) ' = u' v + v ' u and put an integral sign before each term.
It becomes INT (u v) ' = INT (u' v) + INT (v ' u) which is really u v = INT (v du) + INT (u dv) if we just change u' and v' to du and dv.
Now transpose to get u v – INT (v du) = INT (u dv) which is the parts formula.
there are some questions where you may have to make a variable substitution before applying integration by parts, so use w for the change of variable instead of u, so that you can use u for the parts formula.
Watch out for those boomerang integrals that come back to you and smack you in the head.
