Next: About this document ...



SOLUTIONS TO INTEGRATION BY PARTS



SOLUTION 10 : Integrate $ \displaystyle{ \int {x^3 \ln{5x} } \,dx } $ . Let

$ u = \ln{5x} \ \ $ and $ \ \ dv = x^3 dx $

so that

$ du = \displaystyle{1 \over 5x} (5) \ dx = \displaystyle{1 \over x} \ dx \ \ $ and $ \ \ v = \displaystyle{x^4 \over 4} $ .

Therefore,

$ \displaystyle{ \int {x^3 \ln{5x} } \,dx } = \displaystyle{ {x^4 \over 4} \ln{5x} - \int {{x^4 \over 4} } \,{1 \over x} dx } $

$ = \displaystyle{ {x^4 \over 4} \ln{5x} - (1/4)\int {x^3 } \, dx } $

$ = \displaystyle{ {x^4 \over 4} \ln{5x} - (1/4) { x^4 \over 4} + C } $

$ = \displaystyle{ {x^4 \over 4} \ln{5x} - (1/16) x^4 + C } $ .

Click HERE to return to the list of problems.




SOLUTION 11 : Integrate $ \displaystyle{ \int {(\ln x)^2 } \,dx } $. Let

$ u = (\ln x)^2 \ \ $ and $ \ \ dv = dx = (1) dx $

so that

$ du = 2 (\ln{x}) \displaystyle{ 1 \over x } \ dx = \displaystyle{ 2 \ln{x} \over x } \ dx \ \ $ and $ \ \ v = x $ .

Therefore,

$ \displaystyle{ \int {(\ln x)^2 } \,dx } = \displaystyle{ x(\ln x)^2 - \int {x \, {2\ln x \over x} } \,dx } $

$ = \displaystyle{ x(\ln x)^2 - 2\int {\ln x } \,dx } $ .

Use integration by parts again. Let

$ u = \ln x \ \ $ and $ \ \ dv = dx = (1) dx $

so that

$ du = \displaystyle{1\over x} \ dx \ \ $ and $ \ \ v = x $ .

Hence,

$ \displaystyle{ \int {(\ln x)^2 } \,dx } = \displaystyle{ x(\ln x)^2 - 2\int {\ln x } \,dx } $

$ = \displaystyle{ x(\ln x)^2 - 2\Big\{ x \ln{x} - \int { x } \,{1\over x} dx \Big\} } $

$ = \displaystyle{ x (\ln x)^2 - 2x\ln{x} + 2 \int { 1 } \,dx } $

$ = \displaystyle{ x (\ln x)^2 - 2x \ln{x} + 2x + C } $ .

Click HERE to return to the list of problems.




SOLUTION 12 : Integrate $ \displaystyle{ \int { (\ln x)^3 } \,dx } $. Let

ein $ u = ( \ln x)^3 \ \ $ and $ \ \ dv = dx = (1) dx $

so that

$ du = 3(\ln {x})^2 \displaystyle{ 1 \over x } \ dx = \displaystyle{ 3(\ln {x})^2 \over x } \ dx\ \ $ and $ \ \ v = x $ .

Therefore,

$ \displaystyle{ \int{ (\ln x)^3 } \,dx } = \displaystyle{ x(\ln x)^3 - \int { x } \,{ 3(\ln x)^2 \over x} dx } $

$ = \displaystyle{ x(\ln x)^3 - 3 \int { (\ln x)^2 } \, dx } $ .

Use integration by parts again. Let

$ u = (\ln x)^2 \ \ $ and $ \ \ dv = dx = (1) dx $

so that

$ du = 2 (\ln{x}) \displaystyle{ 1 \over x } \ dx = \displaystyle{ 2 \ln{x} \over x } \ dx \ \ $ and $ \ \ v = x $ .

Therefore,

$ \displaystyle{ \int {(\ln x)^3 } \,dx } = \displaystyle{x(\ln x)^3 - 3\int { (\ln x)^2 } \,dx } $

$ = \displaystyle{ x(\ln x)^3 - 3\Big\{ x (\ln x)^2 - \int {x {2\ln {x}\over x} } \,dx \Big\} } $

$ = \displaystyle{ x(\ln x)^3 - 3x (\ln x)^2 + 6\int {\ln x } \,dx } $ .

Use integration by parts for a third time. Let

$ u = \ln x \ \ $ and $ \ \ dv = dx = (1) dx $

so that

$ du = \displaystyle{1\over x} \ dx \ \ $ and $ \ \ v = x $ .

Hence,

$ \displaystyle{ \int {(\ln x)^3 } \,dx } = \displaystyle{ x(\ln x)^3 - 3x (\ln x)^2 + 6\int {\ln x } \,dx } $

$ = \displaystyle{ x(\ln x)^3 - 3x(\ln x)^2 + 6\Big\{ x \ln{x} - \int { x } \,{1\over x} dx \Big\} } $

$ = \displaystyle{ x(\ln x)^3 - 3x(\ln x)^2 + 6x\ln{x} - 6 \int 1 \, dx } $

$ = \displaystyle{ x(\ln x)^3 - 3x(\ln x)^2 + 6 x \ln{x} - 6x + C } $ .

Click HERE to return to the list of problems.




SOLUTION 13 : Integrate $ \displaystyle{ \int {x \sqrt{x + 3} } \, dx } $ . Let

$ u = x \ \ $ and $ \ \ dv = \sqrt {x+3} \ dx = (x+3)^{1/2} \ dx $

so that

$ du = (1) dx = dx \ \ $ and $ v = \displaystyle{ (x+3)^{3/2} \over (3/2) } = (2/3)(x+3)^{3/2} $ .

Therefore,

$ \displaystyle{ \int { x \sqrt{x + 3} } \,dx } = \displaystyle{ x (2/3)(x+3)^{3/2} - \int (2/3)(x+3)^{3/2} \, dx } $

$ = \displaystyle{ (2/3)x(x+3)^{3/2} - (2/3)\int (x+3)^{3/2} \, dx } $

$ = \displaystyle{ (2/3)x(x+3)^{3/2} - (2/3) { (x+3)^{5/2} \over (5/2) }+ C } $

$ = \displaystyle{ (2/3)x(x+3)^{3/2} - (4/15) { (x+3)^{5/2} } + C } $ .

Click HERE to return to the list of problems.




SOLUTION 14 : Integrate $ \displaystyle{ \int { x\sin{x} \cos{x} } \,dx } $ . Let

$ u = x \ \ $ and $ \ \ dv = \sin x \cos x \, dx $

so that

$ du = (1) dx = dx \ \ $ and $ v = (1/2) \sin^2{x} $ .

Therefore,

$ \displaystyle{ \int { x\sin{x} \cos{x} } \,dx } = x (1/2) \sin^2{x} - \displaystyle{ \int (1/2) \sin^2{x} \, dx } $

$ = (1/2)x \sin^2{x} - \displaystyle{ (1/2) \int \sin^2{x} \, dx } $ .

Now use the trig identity

$ \sin^2{x} = (1/2) (1 - \cos 2x ) $ ,

getting

$ \displaystyle{ \int { x \sin{x} \cos{x} } \,dx } = (1/2)x \sin^2{x} - \displaystyle{ (1/2) \int \sin^2{x} \, dx } $

$ = (1/2)x \sin^2{x} - \displaystyle{ (1/2) \int (1/2) (1 - \cos 2x ) \, dx } $

$ = (1/2)x \sin^2{x} - \displaystyle{ (1/4) \int (1 - \cos 2x ) \, dx } $

$ = (1/2)x \sin^2{x} - \displaystyle{ (1/4) \Big\{x - { \sin 2x \over 2 } \Big\} } + C $

$ = (1/2)x \sin^2{x} - (1/4)x + (1/8) \sin 2x + C $ .

Click HERE to return to the list of problems.




SOLUTION 15 : Integrate $ \displaystyle{ \int \Big({ \ln x \over x }\Big)^2 \,dx } = \displaystyle{ \int { (\ln x)^2 \over x^2 } \, dx } $ . Let

$ u = (\ln x)^2 \ \ $ and $ \ \ dv = \displaystyle{ 1 \over x^2} \ dx = x^{-2} \ dx$

so that

$ du = 2 (\ln x) \displaystyle{ 1 \over x } \ dx = \displaystyle{ 2 \ln x \over x } \ dx \ \ $ and $ \ \ v = \displaystyle{ x^{-1} \over -1 } = -\displaystyle{ 1 \over x } $ .

Therefore,

$ \displaystyle{ \int { (\ln x)^2 \over x^2 } \, dx } = \displaystyle{ -{(\ln{x})^2 \over x} - \int { \Big(-{1 \over x} \Big) \,{2\ln x \over x}} \,dx } $

$ = \displaystyle{ -{(\ln{x})^2\over x} + 2 \int{ \ln x \over x^2} \,dx } $ .

Use integration by parts again. Let

$ u = \ln x \ \ $ and $ \ \ dv = \displaystyle{ 1 \over x^2} \ dx $

so that

$ du = \displaystyle{1\over x} \ dx \ \ $ and $ \ \ v = -\displaystyle{ 1 \over x } $ .

Hence

$ \displaystyle{ \int { (\ln x)^2 \over x^2 } \, dx } = \displaystyle{ -{(\ln{x})^2\over x} + 2 \int{ \ln x \over x^2} \, dx } $

$ = \displaystyle{ -{(\ln{x})^2 \over x} + 2 \Big\{ -{1 \over x} \ln{x} - \int{ \Big(-{1\over x} \Big) } \,{ 1 \over x} \, dx \Big\} } $

$ = \displaystyle{ -{(\ln{x})^2 \over x} - { 2\ln{x}\over x } + 2\int{ 1\over x^2} \, dx } $

$ =\displaystyle{ -{(\ln{x})^2\over x} - {2\ln{x}\over x} + 2\Big\{{-1 \over x} \Big\} + C} $

$ =\displaystyle{ -{\ln{x})^2\over x} - { 2\ln{x}\over x} - {2 \over x} + C} $ .

Click HERE to return to the list of problems.




Duane Kouba 2000-04-23