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B
log∣∣∣√x2+2x+4−1√x2+2x+4+1∣∣∣+tan−1(x+23)+C
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C
logtan−1(√x+32)+C
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D
none of these
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Solution
The correct option is D none of these I=∫2x+3(x2+2x+3)√x2+2x+4dx Let I=2I1+I2 Where I1=∫(x+1)dx((x+1)2+2)√(x+1)2+3 Substitute (x+1)2+3=t I1=∫tdt(t2−1)t=12log∣∣∣t−1t+1∣∣∣=12log∣∣∣√x2+2x+4−1√x2+2x+4+1∣∣∣+c And I2=∫dx((x+1)2+2)√(x+1)2+3 Substitute x+1=1t I2=∫(−1t2)t3dt(1+2t2)√1+3t2 Substitute 1+3t=u2 I2=−∫udu(2u2+1)u=−1√2tan−1(√2(x2+2x+4)x+1)