INTEGRACION POR PARTES



 $ \displaystyle{ \int {x e^x} \,dx } $

$ u = x \ \ $   $ \ \ dv = e^x dx $

$ du = (1) dx = dx \ \ $   $ \ \ v = e^x $ .

$ \displaystyle{ \int { x e^x } \,dx } = \displaystyle{ x e^x - \int { e^x} \,dx } $

$ = \displaystyle{ x e^x - e^x + C } $ .



$ \displaystyle{ \int { x \sin {x}} \,dx } $

$ u = x \ \ $      $ \ \ dv = \sin {x} \ dx $

$ du = (1) dx = dx \ \ $     $ \ \ v = -\cos{x} $ .

$ \displaystyle{ \int { x \sin {x}} \,dx } = \displaystyle{ x (-\cos{x}) - \int { (-\cos{x}) } \,dx } $

$ = \displaystyle{ -x \cos{x} + \int { \cos{x} } \,dx } $

$ = -x \cos {x} + \sin{x} + C $
 



 $ \displaystyle{ \int {x \ln{x}} \,dx } $

$ u = \ln{x} \ \ $     $ \ \ dv = x \ dx $

$ du = \displaystyle{1\over x} \ dx \ \ $   $ \ \ v = \displaystyle{x^2 \over 2} $ .

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

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

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

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



$ \displaystyle{ \int { x \cos {3x} } \,dx } $

$ u = x \ \ $   $ \ \ dv = \cos{3x} \ dx $

$ du = (1) dx = dx \ \ $     $ \ \ v = (1/3) \sin{3x}$ .

$ \displaystyle{ \int { x \cos{3x} } \,dx } = \displaystyle{ x (1/3) \sin{3x} - \int { (1/3) \sin{3x} } \, dx } $

$ = \displaystyle{(1/3)x \sin{3x} - (1/3) \int { \sin{3x} } \, dx } $

$ = \displaystyle{ (1/3) x \sin{3x} - (1/3) { -\cos {3x} \over 3 } + C } $

$ =\displaystyle{ (1/3) x \sin{3x} + (1/9) \cos {3x} + C } $ .



 $ \displaystyle{ \int { \ln {x} \over x^5 } \,dx } $

$ u = \ln x \ \ $      $ \ \ dv = \displaystyle{ 1 \over x^5 } \ dx = x^{-5} \ dx $

$ du = \displaystyle{1\over x} \ dx \ \ $       $ \ \ v = \displaystyle{ x^{-4} \over -4} = \displaystyle{ -1 \over 4x^4 } $ .

$ \displaystyle{ \int {\ln {x} \over x^5 } \,dx } = \displaystyle{ (\ln{x})\Big(...

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

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

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

$ = \displaystyle{-{\ln{x} \over 4x^4} - { 1 \over 16x^4} + C } $



 $ \displaystyle{ \int \arcsin 3x \,dx } $

$ u = \arcsin 3x \ \ $                                                 $ \ \ dv = dx = (1) dx $

$ du = \displaystyle{ 1 \over \sqrt{ 1 - (3x)^2 } } (3) dx = \displaystyle{ 3 \over \sqrt{ 1 - 9x^2 } } dx \ \ $          $ \ \ v = x $

$ \displaystyle{ \int \arcsin 3x \, dx } = x \arcsin 3x - \displaystyle{ \int x { 3 \over \sqrt{ 1 - 9x^2 } } \, dx } $

$ = x \arcsin 3x - \displaystyle{ 3 \int {x \over \sqrt{ 1 - 9x^2 } } \, dx } $ .

$ u = 1 - 9x^2 $

$ du = -18x \ dx $ ,

$ (-1/18) du = x \ dx $ .

$ \displaystyle{ \int \arcsin 3x \, dx } = x \arcsin 3x - \displaystyle{ 3\int {x \over \sqrt{ 1 - 9x^2 } } \, dx } $

$ = x \arcsin 3x - \displaystyle{ 3\int {1 \over \sqrt{ 1 - 9x^2 } } \, x \ dx } $

$ = x \arcsin 3x - \displaystyle{ 3\int {1 \over \sqrt{ u }} \, (-1/18) du } $

$ = x \arcsin 3x + \displaystyle{ (1/6) \int {1 \over \sqrt{ u }} \, du } $

$ = x \arcsin 3x + \displaystyle{ (1/6) \int u^{-1/2} \, du } $

$ = x \arcsin 3x + (1/6) \displaystyle{ { u^{1/2} \over (1/2) } } $ + C

$ = x \arcsin 3x + (1/3) \displaystyle{ (1-9x^2)^{1/2} } $ + C

$ = x \arcsin 3x + (1/3) \sqrt{ 1-9x^2 } $ + C



 $ \displaystyle{ \int {\ln x } \,dx } $

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

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

$ \displaystyle{ \int { \ln x } \,dx } = \displaystyle{ x \ln{x} - \int {x {1 \over x}} \,dx } $

$ = \displaystyle{x \ln{x} - \int { 1 } \,dx } $

$ = \displaystyle{ x \ln {x} - x + C } $ .



$ \displaystyle{ \int 2x \arctan x \,dx } $

$ u = \arctan x \ \ $     $ \ \ dv = 2x \ dx $

$ du = \displaystyle{ 1 \over 1 + x^2 } \ dx \ \ $     $ \ \ v = x^2 $ .

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

AGREGANDO $ \ 0 = 1 - 1 \ $  

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

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

$ = x^2 \arctan x - \displaystyle{ \int 1 \, dx

$ = x^2 \arctan x - x + \arctan x + C $ .



 $ \displaystyle{ \int { x^2 e^{3x} } \,dx } $

$ u = x^2 \ \ $    $ \ \ dv = e^{3x} dx $

$ du = 2x \ dx \ \ $      $ \ \ v = (1/3) e^{3x} $ .

$ \displaystyle{ \int { x^2 e^{3x} } \,dx }

$ = \displaystyle{(1/3) x^2 e^{3x} - (2/3)\int { x e^{3x} } \, dx } $ .

$ u = x \ \ $       $ \ \ dv = e^{3x} dx $

$ du = (1) dx = dx \ \ $      $ \ \ v = (1/3) e^{3x} $ .

$ \displaystyle{ \int { x^2 e^{3x} } \,dx } = \displaystyle{ (1/3) x^2 e^{3x} - (2/3)\int { x e^{3x} } \, dx } $

$ = \displaystyle{ (1/3) x^2 e^{3x} - (2/3)\Big\{ x (1/3) e^{3x} -\int { (1/3) e^{3x} } \, dx }\Big\} $

$ = \displaystyle{ (1/3) x^2 e^{3x} - (2/3)\Big\{ (1/3)x e^{3x} - (1/3) \int { e^{3x} } \, dx }\Big\} $

$ = \displaystyle{ (1/3){x^2 e^{3x} } - (2/9) xe^{3x} + (2/9) {e^{3x} \over 3 } + C } $

$ = (1/3){x^2 e^{3x} } - (2/9) xe^{3x} + (2/27) e^{3x} + C $ .