Miscellaneous
Last updated at December 16, 2024 by Teachoo
Transcript
Misc 3 Integrate the function 1/(š„ ā(šš„ ā š„^2 ) ) ā«1ā1/(š„ ā(šš„ ā š„^2 ) ) šš„ = ā«1ā1/(š„ ā(š„^2 (š/š„ ā 1) ) ) šš„ = ā«1ā1/(š„ . š„ā((š/š„ ā 1) ) ) šš„ = ā«1ā1/(š„^2 ā((š/š„ ā 1) ) ) šš„ Let š/š„ ā1=š” Differentiating both sides š¤.š.š”.š„ (ā š)/š„^2 ā0 = šš”/šš„ (ā š)/š„^2 = šš”/šš„ šš„ = (ā š„^2)/š . šš” Putting the values of (a/xā1) and dx, we get ā«1ā1/(š„^2 ā((š/š„ ā 1) ) ) šš„ = ā«1ā1/(š„^2 āš” ) šš„ = ā«1ā1/(š„^2 āš” )Ć(āš„^2)/š . šš” = ā«1ā(ā1)/š šš”/āš” = (ā1)/š ā«1ā(š”)^(ā 1/2) šš” = (ā1)/š [š”^((ā1)/2 + 1)/((ā1)/2 + 1)]+š¶ = (ā1)/š [š”^(1/2)/(1/2)] + š¶ = (ā2)/š [āš”] + š¶ Putting back t = š/š„ā1 = (ā2)/š ā(š/š„ ā1) + š¶ = (āš)/š ā((š ā š)/š) + šŖ