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Please solve the following question

Asked by Balbir 7th October 2017, 11:03 PM
Answered by 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.

end style
Answered by Expert 18th December 2017, 11:21 AM
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