Let x=t6, dx=6t5dt. Then x1/2=t3, x1/6=t, denominator =t(t2+1). Integral =6∫t2+1t4dt=6∫(t2−1+t2+11)dt=2t3−6t+6arctant. Evaluating from t=0 to t=1: 2−6+23π=−4+23π.
If g(x)=∫x1/2+x1/6dx, then g(1)−g(0) is :
Held on 17 Aug 2022 · Verified 13 Jul 2026.
4−23π
5−6loge2
5+6loge2
−4+23π
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