Please solve the following question

Asked by Balbir | 7th Oct, 2017, 11:03: PM

Expert Answer:

begin mathsize 16px style integral fraction numerator straight x to the power of begin display style 1 half end style end exponent over denominator straight x to the power of 1 half end exponent minus straight x to the power of begin display style 1 third end style end exponent end fraction dx
Taking space straight x equals straight t to the power of 6 space as space it space is space LCM space of space 2 space and space 3.
rightwards double arrow dx equals 6 straight t to the power of 5 dt
integral fraction numerator begin display style open parentheses straight t to the power of 6 close parentheses to the power of 1 half end exponent end style over denominator open parentheses straight t to the power of 6 close parentheses to the power of 1 half end exponent begin display style minus end style begin display style begin display style open parentheses straight t to the power of 6 close parentheses end style to the power of 1 third end exponent end style end fraction cross times 6 straight t to the power of 5 dt
integral fraction numerator straight t cubed over denominator straight t cubed minus straight t squared end fraction cross times 6 straight t to the power of 5 dt
integral fraction numerator begin display style straight t cubed end style over denominator begin display style straight t squared open parentheses straight t minus 1 close parentheses end style end fraction cross times 6 straight t to the power of 5 dt
integral fraction numerator 6 straight t to the power of 6 over denominator straight t minus 1 end fraction dt
integral fraction numerator begin display style 6 open square brackets open parentheses straight t cubed close parentheses squared minus 1 plus 1 close square brackets end style over denominator begin display style straight t minus 1 end style end fraction dt
integral fraction numerator 6 open square brackets open parentheses straight t cubed minus 1 close parentheses open parentheses straight t cubed plus 1 close parentheses plus 1 close square brackets over denominator straight t minus 1 end fraction dt
integral fraction numerator begin display style 6 open square brackets open parentheses straight t minus 1 close parentheses open parentheses straight t squared plus straight t plus 1 close parentheses open parentheses straight t cubed plus 1 close parentheses plus 1 close square brackets end style over denominator begin display style straight t minus 1 end style end fraction dt
integral 6 open parentheses fraction numerator open parentheses straight t minus 1 close parentheses open parentheses straight t squared plus straight t plus 1 close parentheses open parentheses straight t cubed plus 1 close parentheses over denominator straight t minus 1 end fraction plus fraction numerator 1 over denominator straight t minus 1 end fraction close parentheses dt
6 integral open parentheses open parentheses straight t squared plus straight t plus 1 close parentheses open parentheses straight t cubed plus 1 close parentheses plus fraction numerator begin display style 1 end style over denominator begin display style straight t minus 1 end style end fraction close parentheses dt
Simplify space the space first space term space and space integrate space using space formulae.

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Answered by Sneha shidid | 18th Dec, 2017, 11:21: AM