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If `int (4e^x+6e^-x)/(9e^x-4e^-x)dx=Ax+B ln (9e^(2x)-4)+C`, then |
Answer» Correct Answer - `A = -(3)/(2), B = (35)/(36) and C in R` Given, `int (4e^(x)+6e^(-x))/(9e^(x) - 4e^(-x))dx = Ax + B log (9e^(2x)-4)+c` `LHS=int(4e^(2x)+6)/(9e^(2x)-4)dx` `Let" "4e^(2x)+6=A(9e^(2x)-4)+B(18e^(2x))` `rArr" "9A + 18B = 4 and -4A = 6` `rArr" "A = -(3)/(2) and B = (35)/(36)` `therefore int(A(9e^(2x)-4)+B(18e^(2x)))/(9e^(2x)-4)dx = A int 1 dx + B int (1)/(t) "dt where t" = 9e^(2x)-4` `=A x + B log(9e^(2x)-4)+c` `=-(3)/(2)x+(35)/(36)log(9e^(2x)-4)+c` `therefore" "A = -(3)/(2), B = (35)/(36)` and c = any real number |
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