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Question : 75 of 160
Marks:
+1,
-0
Solution:
‌‌ Let ‌I=∫‌‌‌ Put ‌z=‌‌⇒‌‌‌=−‌=−z2‌⇒‌‌x+2=‌‌⇒‌‌x=‌‌⇒I=∫‌‌⇒I=−∫‌| dz |
| √(1−2z)2+z(1−2z)+2z2 |
‌=−∫‌| dz |
| √4z2−4z+1+z−2z2+2z2 |
‌=−∫‌‌=−‌‌∫‌ ‌=−‌‌∫‌| dz |
| √z2−‌+‌+‌−‌ |
‌=−‌‌∫‌‌=‌‌ln(z−‌+√(z−‌)2+‌)+C‌=‌‌ln(‌−‌+√(‌−3.‌+‌)+C‌=‌ sinh−1(‌)+C
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