Solve the problem in photo showing each step clearly

Asked by Vidushi412 | 15th Sep, 2019, 11:14: PM

Expert Answer:

begin mathsize 16px style straight I equals integral fraction numerator straight e to the power of straight x open parentheses straight x cubed minus straight x plus 2 close parentheses over denominator open parentheses straight x squared plus 1 close parentheses squared end fraction dx
equals integral open parentheses fraction numerator xe to the power of straight x plus 2 straight e to the power of straight x over denominator straight x squared plus 1 end fraction minus fraction numerator 2 straight x open parentheses xe to the power of straight x plus straight e to the power of straight x close parentheses over denominator open parentheses straight x squared plus 1 close parentheses squared end fraction close parentheses dx
Solve space it space using space integration space by space parts
we space get
equals space fraction numerator 2 open parentheses xe to the power of straight x plus straight e to the power of straight x close parentheses over denominator 2 open parentheses straight x squared plus 1 close parentheses end fraction plus 2 integral open parentheses negative fraction numerator xe to the power of straight x plus 2 straight e to the power of straight x over denominator 2 open parentheses straight x squared plus 1 close parentheses end fraction close parentheses dx plus integral fraction numerator xe to the power of straight x plus 2 straight e to the power of straight x over denominator straight x squared plus 1 end fraction dx
equals fraction numerator xe to the power of straight x plus straight e to the power of straight x over denominator straight x squared plus 1 end fraction plus straight c end style

Answered by Sneha shidid | 16th Sep, 2019, 09:57: AM

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