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Question : 62 of 75
Marks:
+1,
-0
Solution:
‌‌‌dx‌−3‌ln‌√3‌=4‌‌| √3+x2−√1+x2 |
| (3+x2)−(1−x2) |
‌dx−‌‌ln‌3‌=2[{‌√3+x2+‌‌ln(x+√3+x2)}01.‌−{‌√1+x2+‌‌ln(x+√1+x2)}01]−‌‌ln‌3‌=2[{‌√4+‌‌ln(1+√4)}−{0+‌‌ln‌√3}.‌−{‌√2+‌‌ln(1+√2)}+{0+‌(0)}]−‌‌ln‌3‌=2[1+‌‌ln‌3−‌‌ln‌3−‌−‌‌ln(1+√2)]−‌‌ln‌3‌=2+3‌ln‌3−‌‌ln‌3−√2−ln(1+√2)−‌‌ln‌3‌=2−√2−ln(1+√2)
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