MDJH Posted August 1, 2010 Share Posted August 1, 2010 A math professor of mine was recommending that I go over integration techniques before doing courses involving differential equations. Since I plan on doing at least 3 such courses in the fall, (comp. mechanics, classical mechanics, and diff. eq. itself) I need to catch up on this stuff. So, I was doing some integration by parts practice problems, and I've run into trouble already. One such problem was to find the indefinite integral of LN(2x+1)dx with respect to x. So I set up u = LN(2x+1) and dv = dx, implying v = x and du = 2/(2x+1) Using the formula INT(u)dv = uv-INT(v)du, this yielded LN(2x+1)-INT(2/(2x+1))dx So for the integral within an integral, I used parts again. u = 1/(2x+1) and dv = 2dx implies du = -2dx/((2x+1)^2) and v = 2x Using the same formula, this yielded (2x)/(2x+1)-INT((2xdx)/(4x^2+4x+1)) So I tried a w-substitution. If w=4x^2+4x+1, then dw=(8x+4)dx; if it were just 8xdx I could substitute it, but that +4 part makes it seem like it would be mathematically improper. What did I do wrong? Link to comment Share on other sites More sharing options...
D H Posted August 1, 2010 Share Posted August 1, 2010 One such problem was to find the indefinite integral of LN(2x+1)dx with respect to x. So I set up u = LN(2x+1) and dv = dx, implying v = x and du = 2/(2x+1) Correction: [math]du = \frac 2{2x+1} dx[/math]. You forgot the dx. Using the formula INT(u)dv = uv-INT(v)du, this yielded LN(2x+1)-INT(2/(2x+1))dx Look at what you wrote for u, v, and du (with my correction). Do you see the error? Once you make a mistake in an intermediate step all work from that point on is no good. Link to comment Share on other sites More sharing options...
MDJH Posted August 1, 2010 Author Share Posted August 1, 2010 Correction: [math]du = \frac 2{2x+1} dx[/math]. You forgot the dx. Look at what you wrote for u, v, and du (with my correction). Do you see the error? Once you make a mistake in an intermediate step all work from that point on is no good. I'm guessing it was that my initial uv term was supposed to be x*Ln(2x+1) as opposed to just Ln(2x+1) itself? Link to comment Share on other sites More sharing options...
D H Posted August 1, 2010 Share Posted August 1, 2010 That's one error. What about the [math]\int v\,du[/math] term? Link to comment Share on other sites More sharing options...
Srinivasa B Posted August 9, 2010 Share Posted August 9, 2010 A math professor of mine was recommending that I go over integration techniques before doing courses involving differential equations. Since I plan on doing at least 3 such courses in the fall, (comp. mechanics, classical mechanics, and diff. eq. itself) I need to catch up on this stuff. So, I was doing some integration by parts practice problems, and I've run into trouble already. One such problem was to find the indefinite integral of LN(2x+1)dx with respect to x. So I set up u = LN(2x+1) and dv = dx, implying v = x and du = 2/(2x+1) Using the formula INT(u)dv = uv-INT(v)du, this yielded LN(2x+1)-INT(2/(2x+1))dx So for the integral within an integral, I used parts again. u = 1/(2x+1) and dv = 2dx implies du = -2dx/((2x+1)^2) and v = 2x Using the same formula, this yielded (2x)/(2x+1)-INT((2xdx)/(4x^2+4x+1)) So I tried a w-substitution. If w=4x^2+4x+1, then dw=(8x+4)dx; if it were just 8xdx I could substitute it, but that +4 part makes it seem like it would be mathematically improper. What did I do wrong? You have started it right. let, u = ln(2x+1), du = 2.dx/(2x+1) dv = dx, v = x now the integral will become, x.ln(2x+1) - int(2x.dx/(2x+1)) by algebraic manupulation, it will turn out to be, x.ln(2x+1) - int(dx) + int(dx/(2x+1)) which will yield: x.ln(2x+1) - x + ln(2x+1) /2 x.ln(2x+1) - x + ln(2x+1) /2 Verify your result by differentiating again. Link to comment Share on other sites More sharing options...
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