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Asked by Vssanganal | 19 Dec, 2018, 19:21: PM

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begin mathsize 16px style straight I equals integral fraction numerator dx over denominator straight x squared open parentheses straight x to the power of 2009 plus 1 close parentheses to the power of begin display style 2008 over 2009 end style end exponent end fraction Put space fraction numerator root index 2009 of straight x to the power of 2009 plus 1 end root over denominator straight x end fraction equals straight u dx equals fraction numerator 1 over denominator begin display style straight x to the power of 2007 over open parentheses straight x to the power of 2009 plus 1 close parentheses to the power of begin display style 2008 over 2009 end style end exponent end style minus fraction numerator root index 2009 of straight x to the power of 2009 plus 1 end root over denominator straight x squared end fraction end fraction du Solve space it space further space you space will space get straight I equals fraction numerator root index 2009 of straight x to the power of 2009 plus 1 end root over denominator straight x end fraction plus straight c end style

Answered by Sneha shidid | 27 Dec, 2018, 10:32: AM

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