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Question

# Evaluate: ∫dx(x+1)√−x2+x+1 using Euler's substitution.

A
2tan1(t1)+C,t=x2+x+11x
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B
2tan(t1)+C,t=x2+x+11x
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C
2tan1(t+1)+C,t=x2+x+11x
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D
None of these
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Solution

## The correct option is A −2tan−1(t−1)+C,t=√−x2+x+1−1x ∫dx(x+1)(√−x2+x+1) a<0, c>0 ............ Third Euler substitution ⇒x=1−2tt2+1 −x2+x+1=−(1−2t)2+(t2+1)(1−2t)+(t2+1)2(t2+1)2 =−(1+4t2−4t)+t2+1−2t3−2t+t4+1+2t2(t2+1)2 =t4−t2−2t3−2t+1(t2+1)2 =t4−2t3−t2−2t+1(t2+1)2 =(t2−t−1)2(t2+1)2 ⇒√−x2+x+1 x+1=1−2t+1+t2t2+1 =(t−1)2+1t2+1 dn=(−2)(t2+1)−(2t)(1−2t)(t2+1)2dt =−2t2−2−2t+4t2(t2+1)2dt =2t2−2t−2(t2+1)2dt ∫dx(x+1)(√−x2+x+1) 2(t2−t−1(t2+12)dt =∫(t−1)2+1(t2+1)×(t2−t−1)(t2+1) =∫2dt1+(1−t)2 =−2tan−1(t−1)+C √−x2+x+1=xt+1 t=√−x2+x+1−1x

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