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Question : 73 of 160
Marks:
+1,
-0
Solution:
‌∫(1+tan2x)(1+2x‌tan‌x)‌dx=‌∫( sec2x+2x‌tan‌x‌ ‌s‌e‌c2‌x)‌dx=‌∫ sec2x‌dx+∫‌2x‌tan‌x‌ ‌s‌e‌2‌x‌dx=‌tan‌x+2x(‌)−∫tan2x⋅dx+C=‌tan‌x+xtan2x−∫( sec2x−1)‌dx+C=‌tan‌x+xtan2x−tan‌x+x+C=‌x(tan2x+1)+C=x sec2x+C
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