12th Maths 2 | Chapter 3 | Indefinite Integration | Lecture 8 | Exercise 3.3 | Maharashtra Board
12th Maths 2 | Chapter 3 | Indefinite Integration | Lecture 8 | Exercise 3.3 | Maharashtra Board

CBSE Class 12-science Answered

Integration xe2x dx/( 1 + 2x)2 explain in great detail

Asked by haroonrashidgkp | 27 Oct, 2018, 05:40: PM

integral fraction numerator x e to the power of 2 x end exponent over denominator left parenthesis 1 plus 2 x right parenthesis squared end fraction d x l e t space 2 x equals t d i f f e r e n t i a t e space w. r. t space x w e space g e t space 2 d x equals d t d x equals fraction numerator d t over denominator 2 end fraction integral fraction numerator begin display style x e to the power of 2 x end exponent end style over denominator begin display style left parenthesis 1 plus 2 x right parenthesis squared end style end fraction d x equals 1 half integral fraction numerator begin display style t over 2 end style e to the power of t over denominator left parenthesis 1 plus t right parenthesis squared end fraction d t equals 1 fourth integral fraction numerator e to the power of t space t over denominator left parenthesis 1 plus t right parenthesis squared end fraction d t fraction numerator begin display style 1 end style over denominator begin display style 4 end style end fraction integral fraction numerator begin display style e to the power of t left parenthesis space t plus 1 minus 1 right parenthesis end style over denominator begin display style left parenthesis 1 plus t right parenthesis squared end style end fraction d t equals 1 fourth integral e to the power of t left parenthesis fraction numerator t plus 1 over denominator left parenthesis 1 plus t right parenthesis squared end fraction minus fraction numerator 1 over denominator left parenthesis 1 plus t right parenthesis squared end fraction right parenthesis space d t equals 1 fourth integral e to the power of t left parenthesis fraction numerator 1 over denominator 1 plus t end fraction minus fraction numerator begin display style 1 end style over denominator begin display style left parenthesis 1 plus t right parenthesis squared end style end fraction right parenthesis space d t equals 1 fourth space e to the power of t left parenthesis fraction numerator 1 over denominator 1 plus t end fraction right parenthesis plus c equals 1 fourth e to the power of 2 x end exponent left parenthesis fraction numerator 1 over denominator 1 plus 2 x end fraction right parenthesis plus C sin c e space integral e to the power of x left parenthesis f left parenthesis x right parenthesis plus f apostrophe left parenthesis x right parenthesis right parenthesis d x equals space e to the power of x f left parenthesis x right parenthesis space plus C

Answered by Ram Singh | 28 Oct, 2018, 05:36: PM

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