integral of ( xdx)/(x^2+4)(x^2-5)^0.5

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Dear Shivan , &int;(-x dx) /(x 2 + 4)(x 2 -5) 0.5 substitute x 2 -5 = t 2 we get xdx =tdt substitute the two substitutions , &int;-1/(t 2 +9) dt = - tan -1 ( t/3) /3 + c resubstituting the substitution , I = - tan -1 ( (x 2 -5) ^0.5 /3) /3 + c Please feel free to ask your queries here. We are all IITians and here to help you in your IIT JEE preparation. All the best. Win exciting gifts by answering the questions on Discussion Forum. So help discuss any query on askiitians forum and become an Elite Expert League askiitian. Now you score 5+15 POINTS by uploading your Pic and Downloading the Askiitians Toolbar respectively : Click here to download the toolbar.. Askiitians Expert Ajit Singh Verma IITD

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integral of ( xdx)/(x^2+4)(x^2-5)^0.5

integral of (xdx)/(x^2+4)(x^2-5)^0.5

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