Question-208540 Tinku Tara June 17, 2024 None 0 Comments FacebookTweetPin Question Number 208540 by otchereabdullai@gmail.com last updated on 17/Jun/24 Answered by Sutrisno last updated on 18/Jun/24 misal:3x=2tanθx=23tanθ→dx=23sec2θdθ=∫23tanθ−3(2tanθ)2+4.23sec2θdθ=∫23tanθ−34(tan2θ+1).23sec2θdθ=∫23tanθ−32secθ.23sec2θdθ=∫13(23tanθ−3)secθdθ=∫13(23tanθsecθ−3secθ)dθ=13(23secθ−3ln(secθ+tanθ))+c=13(239x2+42−3ln(9x2+42+3x2))+c Answered by Frix last updated on 18/Jun/24 ∫x−39x2+4dx=t=3x+9x3+42∫(19−1t−19t2)dt==t9−lnt+19t=t2+19t−lnt==9x2+49−ln(3x+9x2+4)+C Terms of Service Privacy Policy Contact: info@tinkutara.com FacebookTweetPin Post navigation Previous Previous post: Evaluate-and-leave-your-answer-in-exponent-form-5-2024-5-2023-5-2022-5-2021-5-2020-5-2019-Next Next post: Question-208608 Leave a Reply Cancel replyYour email address will not be published. Required fields are marked *Comment * Name * Save my name, email, and website in this browser for the next time I comment.