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Question Number 55724 by Kunal12588 last updated on 03/Mar/19

Commented by Kunal12588 last updated on 03/Mar/19

$t h i s m a k e m e s t o p l e a r n i n g c a l c u l u s .$
Commented by mr W last updated on 03/Mar/19

$\int \frac{1}{x} d x i s n o t a c e r t a i n v a l u e, b u t f u n c t i o n s .$ $t h e d i f f e r e n c e b e t w e e n a \int \frac{1}{x} d x a n d$ $a n o t h e r \int \frac{1}{x} d x m u s t b e a c o n s t a n t, s o$ $\int \frac{1}{x} d x = 1 + \int \frac{1}{x} d x i s o k .$ $\int \frac{1}{x} d x = 2 + \int \frac{1}{x} d x i s a l s o o k .$ $\int \frac{1}{x} d x = c + \int \frac{1}{x} d x i s a l s o o k .$ $\int \frac{1}{x} d x = c + \int \frac{1}{x} d x = c + \ln x + C = \ln x + C$ $b u t i f y o u c o m p a r e v a l u e s, y o u w i l l$ $g e t$ $\int_{a}^{b} \frac{1}{x} d x = {[\frac{1}{x} x]}_{a}^{b} - \int_{a}^{b} (- \frac{1}{x^{2}}) x d x$ $\int_{a}^{b} \frac{1}{x} d x = {[1]}_{a}^{b} + \int_{a}^{b} \frac{1}{x} d x$ $\int_{a}^{b} \frac{1}{x} d x = 1 - 1 + \int_{a}^{b} \frac{1}{x} d x$ $\int_{a}^{b} \frac{1}{x} d x = \int_{a}^{b} \frac{1}{x} d x$ $t h e r e i s n o c o n t r a d i c t i o n!$ $g e n e r a l l y i t i s n o t w r o n g w h e n y o u w r i t e$ $\int f (x) d x = 1 + \int f (x) d x$
Commented by Kunal12588 last updated on 03/Mar/19

$t h a n k g o d . T h a n k y o u v e r y m u c h s i r .$
Commented by mr W last updated on 03/Mar/19

$g l a d t o s e e t h a t y o u d o c o n t i n u e$ $l e a r n i n g c a l c u l u s . . .$
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