Schaum's Outline of Theory and Problems of Beginning Calculus

~ du=dx306 INTEGRATION BY PARTS [CHAP. 38Integration by parts can be made easier to apply by setting up a box such as the following one forexample (a) above:u=xdo = ex dxU = exIn the first row, we put U and do. In the second row, we write du and U. The result UD - U du isobtained from the box by first multiplying the upper left corner U by the lower right corner o and thensubtracting the integral { D du of the two entries in the second row.Notice that everything depends on a wise choice of U and U. If we had instead picked U = ex anddu = x dx, then U = j x dx = x2/2 and we would have obtainedIwhich is true enough, but of little use in evaluatingintegration 5 xe“ dx by the even more “difficult” integration 1 (x2/2) ex dx.(b) FindIx In x dx. Let us try U = In x and dv = x dx:I u=lnx dv=xdxxex dx. We would have replaced the “difficult”IThen, v = x dx = x2/2. Thus,I IU dV=uv - v dus (c) Find In xLet us try U = In x and L- = dx:r UJX2=-In x -1 {x dx2 2X2 1 x2---lnx---2 22+c= In x dv = dx IIThen, v = dx = x. Thus,I IU dv=ut, - v duIn x dx = x In x - x - dxI I:=xlnx- dxs= x In x - x + C= x(ln x - 1) + C