Integration by parts is used to integrate a product. The formula (given in formula book): One part of the product (u) should be easy to dvdudifferentiate (and will usually simplify when udxuvvdx，？~~dxdxdv(?differentiated). The other part should be easy ；? dx，?The other part One part is to integrate and should not become too much harder is integrated. differentiated when integrated.
xCommon examination questions Definite integrals (using by parts) 2xedxExample 1: Find . ~ Example: a) Find the points where the graph of xxdxlnExample 1: Find . ~This is a suitable candidate for integration by parts ？x yxe，？(2) cuts the x and y axes. dvxThis can be found using integration by parts if we ？xwith : 2anduxe，，yxe，？(2)b) Sketch the graph of . dvdxc) Find the area of the region between the axes and take . uxx，，lnanddu？xdxyxe，？(2)22ux，，，the graph of . dxdu1 ux，，，lndvxxa) Graph cuts y-axis when x = 0, i.e. at y = 2 ，，，evedxx 2dxGraph cuts the x-axis when y = 0, i.e. when x = 2.. dvx，，，xvSubstitute these into the formula: ydx2xxxxxb) The graph looks like: 322222xedxxeedxxeec，？，？？ Substitute these into the formula: ~~ 2.5222 x1x2211？x dxxxxdx，？lnxxdxxlnln，？！22c) Area is . (2)？xedx~~~~1.52xxdxcos2Example 2: Find . x~022111 ，？？xxxcln 24du0.5，？，，？21uxdv Here we take : uxx，，andcosdxx0.511.52 dxdvlnxdxExample 2: Find . ？？xx~，，，？evedu dxux，，，1？？？xxx1lnxdxThis can be thought of as and so can be dx(2)(2)()1？，？？？？！？xedxxeedx~So ~~dv，，，cossinxvxdv？？xx？？？(2)()xeedx = integrated by parts with ux，，lnand1~dxdx？？xxSubstitute these into the formula: ？？？(2)()xee = du1xxdxxxxdxxxxccossinsinsin(cos)，？，？？？ux，，，lnNow that we’ve integrated, we substitute in our ~~ dxx limits: dv，？？xxxcsincos2，，，1vx2？？？xxx ：?(2)(2)？，？？？xedxxeedx~；(0Note: Sometimes it is necessary to use the 01？？2200 ！xdxxxdx，？ln1lnlnxdxxx，？ ，？？？？？02eeeeintegration by parts formula twice (e.g. with ;；;；~~~x2，？，0.13511.135xxdxsin). xxxcln？？~ =