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SCAN0041

# SCAN0041 - i’ To summarize from the beginning we have 1 1...

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Unformatted text preview: i’\ \ To summarize from the beginning, we have 1 1 ——dx= secQ d6: —du=1nu+C. / 7—4 + \$2 ( > / u | | Our answer is in terms of u, but we would like it to be in terms of x: so we need to substitute back: 1 —dil" 11'1’LL“I“O /\/4+x2 H 1n1sec (0) +tan(t9 )| +0 unﬁt—anus) +tan(9 )|+o 1 2 1 1 — — ,. 1n +(2m) +2zt +C' || II In conclusion, +0. dmzln 1 1+12+1 \f—4 + 1:2 V 4:” 23” We can make this answer look nicer: Note that l 1 4 2 1 1-1—1112: 421\$ =§V4+1§2 so 1 1 1 1 1 1+Z\$2+§\$=§V4+\$2+§\$=§(\/4+£B2+ZE) and 1, 1 1 ‘ 1 1n 1—0—1373—1—556 = n 5(x/4+:v2+:v> :ln 5 +111 \/4+iv2+x‘. The 1n|1 /2| can be “absorbed” in the integration constant C. Also, since v4 + 902 + 90 is always positive, we can get rid of the absolute value. This gives us the result 1 _ = 1/ 2 4+\$2d\$ 1n( 4+9: +m)+C 18 ...
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