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Latihan integral parsial • 1. INTEGRAL PARSIAL • 1 6 ( x + 2) 6 6 1 1 ( x + 2) 6 ( x + 2) 6.6 6 6 1 x( x + 2 ) 6 ( x + 2) 7 7 Bila dilanjutnya, bentuk penyelesaiannya menjadi:  7x − x − 2 = ( x + 2) ( x − ( x + 2)) + C = ( x + 2)  6 1 7 6 +C  7 = 1 ( x + 2) 6 (6 x − 2) + C = 7 ( x + 2) 6 (3 x − 1) + C 7 2 • Soal Nomor 2. ∫ 2 x (3 x − 2) 6 dx = u = 2x dv = (3x – 5)6 du= 2 v= 1 (3x − 5)7 21 ∫u dv = ∫ uv − v du 1 1 2 x (3 x − 5) − ∫ (3 x − 5) 7.2 ∫ 2x 7 (3 x − 5) dx = 6 21 21 2x 2 2x 1= (3x − 5) − 7 = (3x − 5) + C 21 8 (3x − 5) − 7 (3x − 5)8 + C 21 21.3.8 252  2x 1  7  2x 3x 5 = (3x − 5)  − 7.(3x − 5)  + C = (3x − 5)  − + +C  21 252   21 252 252   12 x 5  1= (3 x − 5) 7 +  +C = (3x − 5) (12 x + 5) + C 7  252 252  252 • Soal Nomor 3 ∫ 4x 3 x − 2dx = 1 u = 4x dv = (3 x − 2) 2 2 5 du = 4 v= (3 x − 2) 2 9 2 2 8 8 1 2= 4 x (3 x − 2) − ∫ (3 x − 2) 2.4dx 1 3 3 5 2 = x(3 x − 2) −.. Alan Parsons Project Best Torrent. (3 x − 2) 2 + C 2 9 9 9 9 3 5 8 3  2  8 3 15 x − 2(3 x − 2) = (3 x − 2) 2  x − (3 x − 2) + C = (3 x − 2) 2  +C 9  15  9  15  8 3  9 x + 4)  8 3 15 x − 6 x + 4) = (3 x − 2) 2  = (3x − 2) 2  +C 9 15 +C 9  15    8 3 = (3 x − 2) 2 (9 x + 4) + C 135 • dv = ( 3 x − 1) −1 Soal Nomor 4 u = 6x 3 1 −Hasil dari ∫ 6 x(3 x − 1) dx = (UN 2005) 3 du = 6 1 v = ( 3x - 1) 3 2 1 2 − ∫ 6 x(3x −1) dx =6 x. (3 x −1) − ∫ (3 x −1).6dx 2 2 3 1 3 1 3 2 2 1 − 1 3 ∫ 6 x(3x −1) 2 5 dx =3 x(3 x −1) 3 − 3.. (3 x −1) 3 + C 3 3 5 1 − 3 ∫ 2 5 6 x (3 x −1) dx =3 x(3 x −1) − (3 x −1) 3 + C 3 3 5 Jika penyelesaian ini dilanjukan, maka bentuknya menjadi 1 − 3 ∫ 6 x(3x −1) dx =( 3 x −1) (3 x − (3 x −1) + C 2 3 3 5 1 − 2  15 x − 9 x + 3  ∫ 6 x (3 x −1) dx =( 3 x −1) 3  3  +C  15  1 − (6 x + 3) ∫ 6 x(3x −1) dx =( 3 x −1) 2 3 3 +C 15 1 − 1 ∫ 6 x(3x −1) dx = ( 3 x −1) 3 ( 2 x +1) + C 2 3 5 • dv = ( 3- 2 x ) Hasil dari 1 u =x 2∫ x 3 − 2 x dx = (UAN,2004; No.

33) du = dx v =−. Biohazard 1 5 Download Psx Emulator here. 1 2 (3 − 2 x ) 2 3 3 2 = − 1 (3 − 2 x ) 3 2 3∫x 3 − 2 x dx = x.

Solutions to Integration by Partial Fractions Next: SOLUTIONS TO INTEGRATION BY PARTIAL FRACTIONS: Integrate. Factor and decompose into partial fractions, getting (After getting a common denominator, adding fractions, and equating numerators, it follows that; let; let.) (Recall that.).

Click to return to the list of problems.: Integrate. Factor and decompose into partial fractions, getting (After getting a common denominator, adding fractions, and equating numerators, it follows that; let; let.). Click to return to the list of problems.: Integrate. Factor and decompose into partial fractions, getting (After getting a common denominator, adding fractions, and equating numerators, it follows that; let; let.). Click to return to the list of problems.: Integrate. Because the degree of the numerator is not less than the degree of the denominator, we must first do polynomial division. Then factor and decompose into partial fractions, getting (After getting a common denominator, adding fractions, and equating numerators, it follows that; let; let.) (Recall that.).