formula for integration by parts?
403929
Integration by parts is done by breaking down an integral so that\n\nS u dv = uv - S v du (where S is the integration symbol)\n\nFor example, if we were to integrate xcos x dx,\n\nlet u = x\ndu = dx\n\nthen dv = cos x (all the other stuff we didn't use for u)\nv = sin x (just integrate dv)\n\nand the formula will look like\n\nS xcos x dx = xsin x - S sin x dx\n= xsin x + cos x + C\n\nIt's kinda an art form or experience in picking which part of the integral to make u and dv - there's really no rule that says u = cos x.  Experience (or trial and error) will show you that that's probably not the better choice though .
175327
Try this, maybe?\n\nhttp://mathforum.org/library/drmath/drmath.elem.html
917224
Simply use correct order of operations\n\nparentheses\nexponents\nmult/division (from left to right)\naddition/subtraction (from left to right)
527554
Let I=Int((e^5x)*cos 3x)dx               (1)\n\nUse the formula for integration by parts twice\n\nInt(u dv/dx)dx= uv - int(v du/dx)dx\n\ntaking u = e^5x   and  dv/dx=cos 3x\n   du/dx=5(e^5x)  and      v=(1/3)*sin 3x\n\nthen I = (1/3)*e^5x(sin 3x) - (5/3)*int((e^5x)*sin 3x)dx      (2)\n\nUse integration by parts again: -\n\nint((e^5x)*sin 3x)dx = (-1/3)*(e^5x)*(cos 3x)+(5/3)*int((e^5x)*sin 3x)dx\n\nNow the int((e^5x)*sin 3x)dx we have already defined to equal I\n\ntherefore\n\nint((e^5x)*sin 3x)dx = (-1/3)*(e^5x)*(cos 3x)-(5/3)*I\n\nputting this into the equation 2 to get: -\n\nI = (1/3)*e^5x(sin 3x) + (5/9)*(e^5x)*(cos 3x)-(25/9)*I\n\nBring I over to the left hand side: -\n\n(34/9)I = (1/3)*e^5x(sin 3x) + (5/9)*(e^5x)*(cos 3x)\n\nWith a bit of careful rearranging, you can get the answer :-\n\n    I = (1/34)*(e^5x)(3sin3x+5cos3x) + C\n\nwhere C is an arbitrary constant\n\nI know it looks a bit of a bee-hatch written out here, but go through it and make sure you understand what has been done at each step. I may have made a few errors involving minus signs, so make sure you understood what i have done.\n\nI got the answer wrong first time round, but i have gone through it again and corrected the mistakes.
1006943
Rewrite as a power series, since there is no way to do this using traditional methods:\n\nx^2 - (1/2)x^4 + (3/8)x^6 - (37/144)x^8 + (1031/5760)x^10 -.....\n\nThen integrate:\n\n(1/3)x^3 - (1/10)x^5 + (3/56)x^7 - (37/1296)x^9 + (1031/63360)x^11 -.......
760474
simple\n\nintegration = addition \ndifferentiation = division
1065850
(circle's formula) divided by (hexagon's formula)
667880
a) dy/dx = x - 1/x^2\ndy = (x-x^(-2))dx\ny = x^2/2 + 1/x + C\ny(1) = 1^2/2 + 1/1 + C = 2\n1/2 + 1 + C = 2 implies C = 1/2\nAnswer: y = x^2/2 + 1/x + 1/2, x>0\n\nb) The integration is already finished.  Just plug in -1 for x.\ny(-1) = (-1)^2/2 + 1/-1 + C = 1\n-1/2 - 1 + C = 1 implies C = 3/2\nAnswer: x^2/2 + 1/x + 3/2, x<0\n\nc) Just take the derivative of each one.  C1 and C2 are constants, so their derivative is 0.  \n\nd) C1 = 1/2 and C2 = 3/2\n\ne) Follow parts (a) and (b) except plug in 2 for x and set it equal to -1.  Similarly for y(-2) = 2.\nCheck that C1 = -7/2 and C2 = 1/2
829669
integral=4x^-3dx-x^-2dx\n=4(x^-2)/-2-x^-1/-1\n=-2/x^2+1/x+C
1068054
well Y' = dy/dx\nso using seperation of veriables we can get\ndy/dx = K - 1/3*y^3\nand dy/(K-1/3*y^3)=dx now integrate both sides. \nDoes that help?
511329
Start by multiplying both sides by D this gives you\n(D*P) = (2*s*t)\n\nthe divide both sides by 2t which gives you\n(D*P)/(2*t) = s\n\nor\n\ns = (D*P)/(2*t)
1183660
Use integration by parts.\nu = 4x, du = 4dx\ndv = e^-xdx, v = -e^-x\n\nint(udv) = uv - int(vdu), so\nint(4xe^-x)dx = -4xe^-x - int(-e^-x(4dx) + c\n= -4xe^-x + 4int(e^-x)dx + c\n= -4xe^-x - 4e^-x + c\n= -4e^-x(x+1) + c
955693
i don't understand the way you\nhave written down the function\n\nsay that you want to integrate\n2x/(2x-1)\nput u=2x-1,>>>x=(u+1)/2,\ndx=du/2,\nint{2x/(2x-1)}dx\n=int{(2*(u+1)/2)/u}*du/2\n=1/2*int{(u+1)/u}du\n=1/2*int{1+1/u}du\n=1/2[u+lnu]+C\nbut u= 2x-1\n=1/2[2x+ln(2x-1)-1]+C\nwhere C is a constant\n\ni hope that this helps
1156022
Start by letting I = integral of [sin(x)*sin(nx)]\nUse integration by parts twice.\nLet u = sin(x) and dv = sin(nx)\nAfter using integration by parts twice, you'll end up returning to the original integral you started with, so you will have to solve for I.\n(Your instructor should have shown you this trick)\n\nAnswer: [cos(x)*sin(nx) - n*sin(x)*cos(nx)]/[n^2 - 1]\nI did this quickly, so you should check my answer by working it out. \nIf you get stuck, send me an e-mail or add additional details.
933478
Using the Wolfram Integrator:\nhttp://integrals.wolfram.com/index.jsp\n\nOne obtains\n1/10 sec(2x)tan^4(2x) + 2/15 sec(2x)tan^2(2x) + 4/15 tan(2x)\n\nUsing the other poster's logic - you get (u^2 + 1)^3
256029
ummm partial fractions. it depends on what A and B are. basically   you factor the denominator. as much as you can. it should go all the way to   terms that are just (Ax+B)  or  (Cx^2 + Dx + E)     ill assume you get it to this form\nso then 1/ (Ax+B)(Cx^2+Dx+E) = Z / (Ax+B) +    Y/(Cx^2+Dx+E)\nand then multiply   through by ax+b and cx^2+dx+e\nand you get 1 = Z (Cx^2+dx+e)   + Y(ax+b)   and then you expand    1 = zcx^2 + zdx + ze + yax + yb\nsince theres no x^2 or any x terms in the left side  C=0 and D=0 and A =0     so then    ze+yb = 1   and   then u integrate the terms.  you can usually find out all the constants in it and integrate each. the 1/(ax+b) terms integrated is just   ln (ax+b)  and the other terms, theres some equation for. look it up in a integral table. usually u can solve. i dunno.   \n\n\nfor factoring it, u use long division. im gonna assume u know that and if not, let me know somehow. laterz
603820
Well i cannot write the whole formula here but check the link below