Can someone explain this work method. The answer is 16/3 and the problem is to find the following definite integral.

Answer! 5 dx 2x - 1 we can use the substitution u Let 4 = 25-1 du = 2 doc du = dx 2 5 u+ 1 du Utl = 2 20 1 Ju 2 = DC 2 5 utl du 4 5 1 4 . u- 1 / 2 + u-1 / 2 ) du 5 = ) ( 4 112 + 4- 1 12 ) du u 3/2 + 2 4 1 12 7 5 Q0 1 20 312 13 ( 2x - 1 ) + 2 (28-1) 1/2 15 3/ 2 3 4 1 13 / ( 215) - 1) 512 +2 ( 21 5) - 1) "/2 ) - ( 2 ( 21 1 ) - 1 ) + 2 ( 2(1) - 1) 1/2 )4 / 1 3 2 (9 ) 2+ 2 59) -12 1 1) 312 + 21 1)112)) = ) 2 x27 + 6 1 - ( 3 + 2 ) ] 8 -51 - 18+ 6 - 3 24- 2) 8 3 W 8 2 x 8 W 3 16 3 -D Graphically + 4 3 . 2 6 X 2 3 4 5 G