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Q12  Integrate the functions ( x ^3 - 1 ) ^{1/3} x ^ 5

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Given function  ( x ^3 - 1 ) ^{1/3} x ^ 5,

\int ( x ^3 - 1 ) ^{1/3} x ^ 5 dx

Assume the x^3-1 = t  

\therefore 3x^2dx=dt

\implies \int(x^3-1)^{\frac{1}{3}} x^5 dx = \int (x^3-1)^{\frac{1}{3}}x^3.x^2dx

= \int t^{\frac{1}{3}}(t+1)\frac{dt}{3}

= \frac{1}{3} \int \left ( t^\frac{4}{3}+t^\frac{1}{3} \right )dt

 = \frac{1}{3}\left [ \frac{t^{\frac{7}{3}}}{\frac{7}{3}}+\frac{t^{\frac{4}{3}}}{\frac{4}{3}} \right ]+C

= \frac{1}{3}\left [ \frac{3}{7}t^{\frac{7}{3}}+\frac{3}{4}t^{\frac{4}{3}} \right ]+C

= \frac{1}{7}(x^3-1)^{\frac{7}{3}} + \frac{1}{4}(x^3-1)^{\frac{4}{3}} +C , where C is any constant value.

Posted by

Divya Prakash Singh

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