please give me complete solution of 

Asked by Balbir | 23rd Oct, 2017, 07:54: PM

Expert Answer:

begin mathsize 16px style integral straight x cubed over open parentheses 1 plus straight x close parentheses squared dx
Put space 1 plus straight x space equals space straight t space rightwards double arrow space straight x space equals space straight t space minus space 1
rightwards double arrow dx space equals space dt
integral fraction numerator begin display style open parentheses straight t space minus space 1 close parentheses cubed end style over denominator straight t squared end fraction dt
integral fraction numerator straight t cubed minus 3 straight t squared plus 3 straight t minus 1 over denominator straight t squared end fraction dt
integral open parentheses straight t cubed over straight t squared minus fraction numerator 3 straight t squared over denominator straight t squared end fraction plus fraction numerator 3 straight t over denominator straight t squared end fraction minus 1 over straight t squared close parentheses dt
integral open parentheses straight t minus 3 plus 3 over straight t minus fraction numerator begin display style 1 end style over denominator begin display style straight t squared end style end fraction close parentheses dt
Further space you space can space solve space it space using space formulae space of space integration.
end style

Answered by Sneha shidid | 15th Dec, 2017, 10:56: AM