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Question : 70 of 160
Marks:
+1,
-0
Solution:
70. (c)
I=∫dx =3‌∫dx =‌∫dx =‌∫| (7ex+3e−x)−e−x |
| 7ex+3e−x |
dx =‌∫dx−×‌∫dx I=−‌∫dx Put
t=7+3e−2x ⇒dt=−6e−2xdx I=−(−)‌∫ I=‌+‌‌ln|t|+C I=+‌ln(7+3e−2x)+C. ⇒K= and
L= ⇒K+L=+==
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