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Question Number 104197 by mathmax by abdo last updated on 20/Jul/20
calculate∫5+∞dx(x2−9)4
Answered by Dwaipayan Shikari last updated on 20/Jul/20
∫5+∞2xdx2x(x2−9)4=12∫16+∞dt9+t.1t4{t=9tan2θ,1=18tanθsec2θdθdt16∫tan−143π218tanθsec2θdθsecθ.94tan8θ{takex2−9=t13.193∫tan−143π2secθtan7θdθ.....continue
Answered by mathmax by abdo last updated on 20/Jul/20
A=∫5∞dx(x2−9)4⇒A=∫5∞dx(x−3)4(x+3)4=∫5∞dx(x−3x+3)4(x+3)8wedothechangementx−3x+3=t⇒x−3=tx+3t⇒(1−t)x=3t+3⇒x=3t+31−tdxdx=3(1−t)+(3t+3)(1−t)2=6(1−t)2x+3=3t+31−t+3=3t+3+3−3t1−t=61−t⇒A=∫1416(1−t)2t4(61−t)8dt=167∫141(t−1)8(t−1)2t4dt=167∫141(t−1)6t4dt=167∫141∑k=06C6ktkt4dt=167∑k=06C6k∫141tk−4dt=167∑k=0andk≠36C6k[1k−3tk−3]141+167C63[lnt]141A=167∑k=0andk≠36C6kk−3{1−14k−3}+C6367×(2ln2)
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