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Question : 75 of 160
Marks:
+1,
-0
Solution:
Given,
∫(x+2)√x2−x+2‌dx ⇒‌‌‌f(x)+‌g(x)+‌h(x)+C Let
x+2=A(2x−1)+B Comparing coefficients, we get
‌2A=1 ⇒A=‌,−A+B=2 ⇒B=2+A=2+‌=‌ So,
x+2=‌(2x−1)+‌ ‌∫(x+2)√x2−x+2‌dx ‌=∫(‌(2x−1)+‌)√x2−x+2‌dx ‌=‌‌∫(2x−1)√x2−x+2‌dx+‌‌∫√x2−x+2‌dx Let
u=x2−x+2, then
du=(2x−1)‌dx ‌=‌‌∫√udu+‌‌∫√(x−‌)2+(‌)2‌dx ‌=‌⋅‌+‌[‌√x2−x+2+‌‌ln‌|x−‌+√x2−x+2|]+c ‌=‌(x2−x+2)‌+‌√x2−x+2+‌‌ln‌|x−‌+√x2−x+2|+c Comparing with original equation, we get
‌f(x)=(x2−x+2)‌ ‌g(x)=(2x−1)√x2−x+2 ‌h(x)=ln‌|x−‌+√x2−x+2| So,
f(−1)=[(−1)2−(−1)+2]‌ ‌=[1+1+2]‌=4‌=8 ‌g(−1)=(2(−1)−1)√(−1)2−(−1)+2 ‌=−3√1+1+2=−3√4 ‌=−3×2=−6‌h(‌)=ln‌|‌−‌+√(‌)2−‌+2|‌⇒ln‌|0+√‌−‌+2|‌⇒ln‌|√‌|⇒ln‌|√‌|=ln(‌)‌‌ So, ‌f(−1)+g(−1)+h(‌)=8−6+ln(‌)‌=2+ln(‌)
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