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The value of dxx+x3 is 

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a
3x+3(x3)−6x6+6log⁡(x6+1)+C
b
2x+6(x6)−6log⁡(x6+1)+C
c
2x−3(x3)+6(x6)−6log⁡(x6+1)+C
d
None of the above

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detailed solution

Correct option is C

Let ⁡I=∫dxx+x3Now let  x=t6⇒dx=6t5dt⇒ I=∫6t0t3+t2dt=6∫t3t+1dt=6∫t2−t+1−1t+1dt=2t3−3t2+6t−6log⁡(t+1)+C=2x−3(x3)+6(x6)−6log⁡(x6+1)+C


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