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Question : 123 of 160
Marks:
+1,
-0
Solution:
‌∫(1+x−x−1)ex+x−1‌dx‌=∫ex+x−1‌dx+∫(x−‌)ex+x−1‌dx‌=ex+x−1‌∫dx−∫[‌(ex+x−1)]x‌dx‌+∫(x−‌)ex+x−1‌dx‌=xex+x−1−∫(1−‌)xex+x−1‌dx‌+∫(x−‌)ex+x−1‌dx‌=xex+x−1−∫(x−‌)ex+x−1‌dx‌+∫(x−‌)ex+x−1‌dx+C‌=xex+x−1+C‌‌l
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