answer-to-25824-we-have-a-x-2-e-x-2-ln-a-so-for-a-gt-1-ln-a-ln-a-1-2-2-gt-gt-gt-gt-R-a-x-2-R-e-x-ln-a-1-2-2-dx-and-with-the-changement-t-x-ln-a- Tinku Tara June 4, 2023 Integration 0 Comments FacebookTweetPin Question Number 25932 by abdo imad last updated on 16/Dec/17 answerto25824wehavea−x2=e−x2ln(a)sofora>1ln(a)=((ln(a))1/2)2>>>>∫Ra−x2=∫Re−(x(ln(a)1/2)2dxandwiththechangementt=x(ln(a)1/2>>>>x=t(ln(a))−1/2wehave∫Ra−x2dx=π1/2(ln(a))−1/2…if0<a<1ln(a)<0andtheintegraleisdivergente… Terms of Service Privacy Policy Contact: info@tinkutara.com FacebookTweetPin Post navigation Previous Previous post: A-line-passes-through-A-3-0-and-B-0-4-A-variable-line-perpendicular-to-AB-is-drawn-to-cut-x-and-y-axes-at-M-and-N-Find-the-locus-of-the-point-of-intersection-of-the-lines-AN-and-BM-Next Next post: 4-27-4-29-4-31-4- Leave a Reply Cancel replyYour email address will not be published.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.