Problemas resueltos de integral indefinida resueltos de integral indefinida Matemticas I Curso 2012-2013 222 sen x cos x edx senx cosx + = sen x cos x sen x 2cos x senx 2cosx22 2 22 senx cosx senx cosx senx cosx cosx senx + =+ =+ =

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Problemas resueltos de integral indefinida Matemticas I Curso 2012-2013 14. Calcular las siguientes integrales 3( ) ( )a Ln x dx =( ) ( )3 23133u Ln x du Ln x dxLn x x xxdv dx v x = = = = = ( )2 1Ln xxpor partesdx = ( ) ( ) ( )23 212 ln3u Ln x du x dxLn x x Ln x x xxdv dx v x = = = = = = ( ) 12 ln xxdx = ( ) ( )3 213 6 lnpor partesu Ln x du dxLn x x Ln x x x dx xdv dx v x = = = + = = = = ( ) ( ) ( ) ( ) ( ) ( )3 2 3 213 6 3 6 6Ln x x Ln x x Ln x x x dx Ln x x Ln x x Ln x x x Cx = + = + + 2 23 3 1( ) 11 1xx xxt e x Lnte e t t tb dx dte t tdx dtt = = = = = + += (1 3 ) 11tt t+(1 3 )1tdt dtt= =+ (1 3 ) 4 13 3 43 1 41 1 131 1D rD d c r ct d ddt dt dt dttt t tt t = + = + = = = + = + + + + + = + + + ( )43 4 1 3 4 1 3 1x x x xt Ln t C e Ln e C e Ln e C= + + + = + + + = + + + ( )1/ 2 6 1/ 254/3 3/ 2 6 4/3 6 3/ 26 53. . .(2,3) 6 ( )( ) 6( ) ( )6m c mx x tc dx dx t dtx x t tx t dx t dtx x x = = = = = + += =+ 3 85 68 9 8 916 6 6 6 1 6 11t tt dt dt dt Ln t C Ln x Ct t t t t= = = = + + = + ++ + + 2( )( )2Ln x Lntx t x td dxtx dx t dt = = = = = 2 t12por partesu Lnt du dtdt Lnt dt tdv dt v t = = = = = = = 12 2 2 1 2 2 2 2t Lnt t dt t Lnt dt t Lnt t C x Ln x x Ct = = = + = + Problemas resueltos de integral indefinida Matemticas I Curso 2012-2013 2 22( ) sen x cos xe dxsen x cos x+=2 2 2 22sen x cos x sen x 2cos x sen x 2cos xsen x cos x sen x cos x sen x cos x cos x sen x+= + = + = 2 2 2sen x cos x sen x cos xdx dx dx dx Ln cos x Ln sen x Ccos x sen x cos x sen x= + = + = + + 2( ) 4f x dx 2224 4 1 2 14 2x xx dx dx dx = = ( )2 222 2 1 2 42x sen t dx cost dtsen t cost dt cos t dtxt=arc sen = = = = = = 2 1 (2 ) 1 (2 )4 2 1 2 (2 ) 2 (2 )2 2cos t cos tcos t dt dt cos t dt t sen t C + + = = = = + = + + = 22 2x xarc sen sen 2arc sen C = + + 3 3 33 3(2 ) 2 (2 )1 2( ) (2 ) (2 )( ) (2 )1 3 3( )3x x xx xpor partesu sen x du cos x dxg e sen x dx sen x e e cos x dxdv e dx v e = = = = = = = ( )3 3 33 3(2 ) 2 (2 )1 2 1 1(2 )( ) (2 )( ) 2 ( ) (2 )1 3 3 3 3( )3x x xx xu cos x du sen x dxsen x e cos x e e sen x dxdv e dx v e = = = = + = = Si llamamos I a la integral original, tenemos: 3 31 2 4(2 ) (2 )3 9 9x xI sen x e cos x e I= ( )3 3 3913 1 2 1 2(2 ) (2 ) (2 ) (2 )9 3 9 3 913x x xI sen x e cos x e I e sen x cos x C= = +Problemas resueltos de integral indefinida Matemticas I Curso 2012-2013 ( )( ) ( ) ( )( ) ( )2 2 2223 522 2 2 2 423 51 1 1 1( ) 1 2121 2 1 2 21 3 52 21 13 5x x xx xx xe t e t e t x Ln th e e dx tdx dttt t tt t dt t t dt t dt t dt Cte e C = = = = = = = = = = = + = = + + 21 2(1 )( )x xx xIIe x Ln x ei dx dx e Ln x dx e Ln x Ix x+= + = ( ) 2I+ xe Ln x C= + 1 221x xx x xIu e du e dxI e Ln x e Ln x dx e Ln x Idv dx v Ln xx = = = = = = = ( )1( ) 11 1xxe t x Lntdxj dte t tdx dtt = = = = + += ( ) ( ) ( ){1 1 (1 ) 1 (1 ) 1 11 1 1 1A B A t Bt A t Bt A Bt t t t t t t t+ += + = = + + = =+ + + + ( )( )1 1 1 1 11 1 1x xxdx dt dt Ln t Ln t C Ln e Ln e Ce t t t t = = = + = + + + + + 3 32 3 33211 1 1( )3 3 3 93u Ln x du dxx xxk x Ln x dx Ln x x dx Ln x x Cxxdv x dx v = = = = = + = = ( ) ( ) ( )( )2 2 22 22 2 2 2 2 21 1( ) 1 2 221 1 1 1 1 1122 2 2 2 2 4x x xx xx x x x x xu x du dxl x e dx x e e dxdv e dx v exe e dx xe e C xe e C = = = = = = = = + = + + = +

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