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Mathematics 8 Online
OpenStudy (anonymous):

Hi, Could someone walk me through this question: Evaluate the improper integral sinh2x/1+cosh^4x dx between 0 and infinity. So far I have changed sinh2x to 2sinhxcoshx and I think I will use u substitution and make u=coshx Also du=sinhx, so I have u.du/1+u^4...does that mean I integrate to get (u^2/2) / x+u^5/5 ? And then I get to 5u^2/x+u^5... Then for x i have cosh-1u? So overall it is 5u^2/cosh-1u+u^5

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