.. 98 Section.6 Integration by Parts Integration by parts is another technique that we can use to integrate problems. Typically, we save integration by parts as a last resort when substitution will not work. To see where the integration by parts formula comes from, let's examine the product rule for differentiation and then integrate both sides: " d dx [st nd ] st Derivative + nd Derivative " of the nd of the st d dv du [u v] u + v dx dx dx Now, integrate both sides with respect to x: d dx [u v] dx dv du (u + v dx dx )dx Integrate term by term on the right side: d dx [u v] dx dv u dx dx + du v dx dx Pretend that dv du and dx dx d dx [u v] dx u dv + v du Integrate the left side: u v u dv + v du Now subtract v du from both sides: u v v du u dv are fractions & reduce the "dx". Finally, switch the sides to give us the formula: Integration by Parts formula: u dv u v v du In using this formula, we need to examine what we are integrating and identify a function and a derivative. We then differentiate the function and integrate the derivative. Finally, we plug into the Integration by Parts formula and finish the problem. Let's outline the steps:
Integration by Parts: I) Identify u and dv. Find du by differentiating u and find v by integrating dv. When finding v, do not worry about + c. Plug our results into u dv u v v du & integrate. Note if v du is worse than the integral we started with, we made the wrong choice for u and dv in step #I. When integrating in general, one should follow these steps: ) Check to see if we can integrate using one of the Rules of Integration from Sect.. ) If that does not work, see if we can integrate by Substitution. ) If that fails, then we will use Integration by Parts. Integrate the following: Ex. xex dx. integration from sect.. Also, substitution will not help either. Thus, we need to use integration by parts: I) Let u x and dv e x dx. Then du dx and v ex dx ex. Plug into the integration by parts formula & integrate: u dv u v v du xex dx x ex ex xex dx ex dx xex ex + c xex ex 9 + c. Notice that we use + c at the end of the problem since we have an indefinite integral. We do not need to use + c when find v since we take care of it at the end... 99
.. 00 Ex. 4t ln (t) dt. First, find the indefinite integral. This problem does not fall into one of our rules of integration from sect.. Also, substitution will not help either. Thus, we need to use integration by parts: I) Let u ln (t) and dv 4t dt. Then du t dt and v 4t dt 4t t. Plug into the integration by parts formula and integrate: u dv u v v du 4t ln (t) dt ln (t) t t dt t t ln (t) t dt t ln (t) t + c. Thus, 4t ln (t) dt t ln (t) t [() ln() () ] [() ln() () ] [8 ln() 4] [0 ] 8 ln().4 Ex. e 4x dx Note: e 4x dx e 4x dx. Since this falls under one of our rules of integration from Section. (#4 ekx dx ekx k integrate: e 4x dx 4x e 4 + c, k 0), we will use that to + c.e 4x + c.
.. 0 Ex. 4 0 x e x dx First, find the indefinite integral. This problem does not fall into one of our rules of integration from sect.. Also, substitution will not help either. Thus, we need to use integration by parts: We would like to use e x dx as our dv, but we are missing a factor of x since the derivative of x x. But, we can rewrite the integral as: x e x dx x (x e x )dx x (x e x )dx I) Let u x and dv x e x dx Then du x dx and v x e x dx. To find v, we will need to use substitution: Let u* x, then du* x dx. Substituting, we get: v x e x dx eu* du* e u* + c. Replacing u* by x, we obtain: v e x (ignore the + c for now). Plug into the integration by parts formula & integrate: u dv u v v du x (x e x )dx [x e x ex x dx ] But, from above, ex x dx e x + c, Thus, [x e x ex x dx ] [x e x e x ] + c Thus, 0 x e x ex + c. x e x dx x e x ex 0
.. 0 () ( e () ( e() (0) ) ( e (0) e e ) (0 ) e(0) ) Ex. x x dx integration from sect.. However, we can use substitution to work this problem: Let u x and u + x (add to both sides) du dx Substituting in, we get: x x dx (u + ) u du (u + ) u/ du (u/ + u / ) du u/ + u/ + c u + u + c Replacing u by x, we obtain: ( x ) + ( x ) + c. Ex. 6 Find x x dx using integration by parts Now, we are being forced to use integration by parts: I) Let u x and dv x dx (x ) / dx Then du dx and v (x )/ dx. To find v, we will use substitution: Let u* (x ) du* dx Substituting, we find: v (x )/ dx u*/ du* u*/ Replacing u* by (x ) yields: + c u*/ + c.
.. 0 v (x )/ (ignore the + c for now) Plug into the integration by parts formula & integrate: u dv u v v du x x dx x (x )/ (x )/ dx x(x )/ (x )/ dx To find (x )/ dx, we will need to use substitution again: Let u** (x ) du** dx Substituting yields: v (x )/ dx u**/ du** u**/ u**/ + c. Replacing u** by (x ) yields: (x )/ (ignore the + c for now) + c Hence, x(x )/ (x )/ dx x(x )/ (x )/ + c x(x )/ 4 (x )/ + c x ( x ) 4 ( x ) + c Notice that the answer appears to be different from the answer in Example # ( ( x ) + ( x ) + c). These two answers are actually equivalent. To see how, consider: Ex. 7 Show that the answers from Example and 6 are equal. We need to show that: x x ) ( 4 ( x ) ( x ) + Build each side up so it has a denominator of : ( x )
.. 04 x (x ) 4 ) 0x ) 4 Factor out ) [0x 4 But (x ) 6 ) (x ) + 0 + ) ( x ) from the numerators: (x ) ] ) [6 (x ) + 0] (x ) ( x ) x x since x 0 (for the original problem x x dx to be defined, x has to be greater than or equal to 0). ) [0x 4(x )] ) [0x 4x+ ] ) [6x+] ) [6(x ) + 0] ) [6x 8+ 0] ) [6x+] Hence, they are equal. Given the choice of substitution or integration by parts, many times it is easier to use substitution as Examples and 6 illustrate. That's why we use integration by parts as a last resort. Integrate the following: x x Ex. 8 dx If we go ahead and divide, we can then integrate using our rules of integration from sect.: x x dx ( x ) dx dx x dx x ln ( x ) [ ln()] [ ln()] [ ln()] [ 0] 4 ln() 0.88.
.. 0 Ex. 9 x e x dx integration from sect.. Also, substitution will not help either. Thus, we need to use integration by parts: I) Let u x and dv e x dx. Then du x dx and v ex dx e x. Plug into the integration by parts formula & integrate: u dv u v v du x e x dx x e x ex x dx x e x [ xex dx] To integrate xex dx, we need to use integration by parts again: I) Let u* x and dv* e x dx. Then du* dx and v* ex dx e x. Plug into the integration by parts formula & integrate: u* dv* u* v* v* du* xex dx xe x ex dx xe x e x. (Ignore + c for now). Thus, x e x [ xex dx] x e x [xe x e x ] + c x e x xe x + e x + c.
Ex. 0 x x+ dx integration from sect.. However, we can use substitution to work this problem: Let u x + and u x (Subtract from both sides) du dx, solving for dx yields: du dx Substituting in, we get: x x+ dx u u du u u du u du u ln ( u ) + c. du ( u ) du Replacing u by x + yields: u ln ( u ) + c (x + ) ln ( x + ) + c x + ln ( x + ) + c... 06 Ex. ln (x) dx integration from sect.. Also, substitution will not help either. Thus, we need to use integration by parts: I) Let u ln (x) and dv dx. Then du x dx and v dx x. Plug into the integration by parts formula and integrate: u dv u v v du ln (x) dx x ln (x) x x dx x ln (x) dx x ln (x) x + c.