Question-131288 Tinku Tara June 3, 2023 None FacebookTweetPin Question Number 131288 by mohammad17 last updated on 03/Feb/21 Answered by liberty last updated on 11/Feb/21 L=∫1−3x−3x8dx=∫x−41−3x−3dxchangeofvariable:1−3x−3=hor1−3x−3=h2⇒9x−4dx=2hdhL=29∫h2dh=227h3+cL=227(x3−3x3)3+c=227(x3−3x3)x3−3x3+c Answered by Dwaipayan Shikari last updated on 03/Feb/21 ∫x3−3x11dx=∫1x41−3x3dx=19∫9x41−3x3dx=227(1−3x3)32+C Terms of Service Privacy Policy Contact: info@tinkutara.com FacebookTweetPin Post navigation Previous Previous post: evaluate-f-x-dx-where-f-x-e-x-x-0-1-x-0-lt-x-1-1-x-2-1-lt-x-2-5-2-lt-x-5-5-1-x-5-2-x-gt-5-Next Next post: 1-1-tan-x-arctan-x-dx-