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CBSE Class 12-science Answered

Please solve the following
Asked by Balbir | 06 Oct, 2017, 11:37: AM
answered-by-expert Expert Answer
begin mathsize 16px style integral fraction numerator dx over denominator open parentheses straight x plus 2 close parentheses square root of straight x squared plus 6 straight x plus 7 end root end fraction Put space log space open parentheses straight x plus 2 close parentheses equals straight t space rightwards double arrow fraction numerator 1 over denominator straight x plus 2 end fraction dx equals dt rightwards double arrow straight x plus 2 equals straight e to the power of straight t rightwards double arrow straight x equals straight e to the power of straight t space minus space 2 integral fraction numerator begin display style fraction numerator dx over denominator straight x plus 2 end fraction end style over denominator square root of straight x squared plus 6 straight x plus 7 end root end fraction integral fraction numerator dt over denominator square root of begin display style open parentheses straight e to the power of straight t space bold minus bold space 2 close parentheses squared plus 6 open parentheses straight e to the power of straight t space minus space 2 close parentheses plus 7 end style end root end fraction integral fraction numerator dt over denominator square root of straight e to the power of 2 straight t end exponent minus 4 straight e to the power of straight t plus 4 plus 6 straight e to the power of straight t minus 12 plus 7 end root end fraction integral fraction numerator dt over denominator square root of straight e to the power of 2 straight t end exponent plus 2 straight e to the power of straight t minus 1 end root end fraction integral fraction numerator begin display style dt end style over denominator begin display style square root of straight e to the power of 2 straight t end exponent plus 2 straight e to the power of straight t plus 1 minus 2 end root end style end fraction integral fraction numerator dt over denominator square root of begin display style open parentheses straight e to the power of straight t plus 1 close parentheses squared minus open parentheses square root of 2 close parentheses squared end style end root end fraction Use space formula space for space the space integration space and space solve space it space further. end style
Answered by Sneha shidid | 22 Dec, 2017, 09:39: AM

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