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Question : 70 of 120
Marks:
+1,
-0
Solution:
Let
I=∫‌dx =∫‌dx Let
ex=t exdx=dt dx=‌dt ∴‌‌I=∫‌×‌dt I=∫‌dt....(i)
Applying partial fraction,
‌=‌+‌+‌ Comparing the coefficient of
t2,t and constant,
1=t2(A+B+C)+t(B−C)+(−A) −A=1⇒A=−1 B−C=0⇒B=C A+B+C=0 −1+2B=0 B=‌ and
C=‌ =−log|t|+‌‌log|t2−1|+C =−log‌ex+‌‌log|e2x−1|+C =−x+‌‌log|e2x−1|+C Hence,
I=‌‌log|e2x−1|−x+C
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