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The constant of integration is missing in the first equation.This falls under the trick "0 = 0", i.e.d(1+C) = d(1); C is any constant.
try a definite integral from 3 to 4 let me have your ans..
int^{4}_{3} x*d(1/x) + int^{4}_{3} (1/x)*d(x) = int^{4}_{3} d(x*(1/x))=>int^{4}_{3} x*(-1/x^2) + int^{4}_{3} (1/x)*1 = int^{4}_{3} d(1)=>int^{4}_{3} (-1/x) + int^{4}_{3} (1/x) = 1|_{4} - 1|_{3}=> 0 = 0.A constant (here 1) evaluated at 4 or 3 remains intact in value (thats why it is a constant).-Prashanth.
Differentiation of a constant is zero
=uv=\int udv+\int vdu} "{\color{black} \int d(uv)=uv=\int udv+\int vdu}")
dx+\int \frac{1}{x}dx} "{\color{black} 1=\int x.(\frac{-1}{x^{2}})dx+\int \frac{1}{x}dx}")
The constant of integration is missing in the first equation.
ReplyDeleteThis falls under the trick "0 = 0", i.e.
d(1+C) = d(1); C is any constant.
try a definite integral from 3 to 4 let me have your ans..
Deleteint^{4}_{3} x*d(1/x) + int^{4}_{3} (1/x)*d(x) = int^{4}_{3} d(x*(1/x))
Delete=>
int^{4}_{3} x*(-1/x^2) + int^{4}_{3} (1/x)*1 = int^{4}_{3} d(1)
=>
int^{4}_{3} (-1/x) + int^{4}_{3} (1/x) = 1|_{4} - 1|_{3}
=> 0 = 0.
A constant (here 1) evaluated at 4 or 3 remains intact in value (thats why it is a constant).
-Prashanth.
Differentiation of a constant is zero
ReplyDelete