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

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$

1136 Views

Integrals

Hope you found this question and answer to be good. Find many more questions on Integrals with answers for your assignments and practice.

Mathematics Part II

Browse through more topics from Mathematics Part II for questions and snapshot.