The value of ∫dxx+x3 is

The value of dxx+x3 is

  1. A

    3x+3(x3)6(x6)+6log(x6+1)+C

  2. B

    2x+6x66log(x6+1)+C

  3. C

    2x3(x3)+6(x6)6log(x6+1)+C

  4. D

    none of these

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    Solution:

    We want a substitution that will allow us to 

    find both the square root and the cube root without getting 

    fractional exponents. Thus we want a substitution of the 

    form x=uk where k is a multiple of 2 and 3. Let us use the
    least common multiple 6.

    Let x=u6,dx=6u5du

    dxx+x3=6u5duu3+u2=6u3u+1du=6u2u+11u+1du=2u33u2+6u6log(u+1)+C=2x3(x3)+6(x6)6log(x6+1)+C

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