+91-8076753736[email protected]
logo
logo
-->

Integrals

Question
CBSEENMA12035638
Wired Faculty App

space F i n d space colon integral fraction numerator left parenthesis 2 x minus 5 right parenthesis e to the power of 2 x end exponent over denominator left parenthesis 2 x minus 3 right parenthesis end fraction d x

Solution

Consider the given integral
straight I space equals integral fraction numerator left parenthesis 2 straight x minus 5 right parenthesis straight e to the power of 2 straight x end exponent over denominator left parenthesis 2 straight x minus 3 right parenthesis cubed end fraction dx Rewriting space the space above space integral space as straight I space equals space integral straight e to the power of 2 straight x minus 3 end exponent space straight x space straight e cubed fraction numerator left parenthesis 2 straight x minus 3 minus 2 right parenthesis over denominator left parenthesis 2 straight x minus 3 right parenthesis cubed end fraction dx equals space straight e cubed integral straight e to the power of 2 straight x minus 3 end exponent open square brackets fraction numerator left parenthesis 2 straight x minus 3 right parenthesis over denominator left parenthesis 2 straight x minus 3 right parenthesis cubed end fraction minus fraction numerator 2 over denominator left parenthesis 2 straight x minus 3 right parenthesis cubed end fraction close square brackets dx equals space straight e cubed integral straight e to the power of 2 straight x minus 3 end exponent open square brackets fraction numerator 1 over denominator left parenthesis 2 straight x minus 3 right parenthesis squared end fraction minus fraction numerator 2 over denominator left parenthesis 2 straight x minus 3 right parenthesis cubed end fraction close square brackets dx Let space us space consider comma space 2 straight x minus 3 space equals space straight t rightwards double arrow space 2 dx space equals space dt therefore space straight I equals straight e cubed integral straight e to the power of straight t open square brackets 1 over straight t squared minus 2 over straight t cubed close square brackets dt over 2 rightwards double arrow space straight I equals straight e cubed over 2 integral straight e to the power of straight t open square brackets fraction numerator straight t minus 2 over denominator straight t cubed end fraction close square brackets space dt Let space straight f space left parenthesis straight t right parenthesis space equals 1 over straight t squared straight f apostrophe left parenthesis straight t right parenthesis space equals space fraction numerator negative 2 over denominator straight t cubed end fraction If space straight I space equals integral straight e to the power of straight t space end exponent left square bracket straight f space left parenthesis straight t right parenthesis space plus space straight f apostrophe left parenthesis straight t right parenthesis space right square bracket dt space then comma space straight I space equals space space straight e apostrophe straight f space left parenthesis straight t right parenthesis space plus space straight C therefore space straight I space equals straight e cubed over 2 space straight x space straight e to the power of straight t space space straight x space straight f space left parenthesis straight t right parenthesis space plus space straight C equals space straight e cubed over 2 space straight x space space straight e to the power of 2 straight x minus 3 end exponent space space straight x space fraction numerator 1 over denominator left parenthesis 2 straight x minus 3 right parenthesis squared end fraction space plus space straight C equals fraction numerator straight e to the power of 2 straight x end exponent over denominator 2 left parenthesis 2 straight x minus 3 right parenthesis squared end fraction plus straight C

Some More Questions From Integrals Chapter

Mock Test Series

Mock Test Series

NCERT Sample Papers

Entrance Exams Preparation