Mathematics
##### ASSERTION

STATEMENT-1 : $\displaystyle \int \frac{\left \{ f(x)\phi '(x)-f'(x)\phi (x) \right \}}{f(x)\phi (x)}\left \{ \log \phi (x)-\log f(x) \right \}dx=\frac{1}{2}\left \{ \log \frac{\phi(x)}{f(x)} \right \}^{2}+c$

##### REASON

STATEMENT-2 : $\displaystyle \int (h(x))^{n}h'(x)dx=\frac{(h(x))^{n+1}}{n+1}+c$

##### ANSWER

STATEMENT-1 is True, STATEMENT-2 is True; STATEMENT-2 is a correct explanation for STATEMENT-1

##### SOLUTION
$\displaystyle I=\int \frac{\left \{ f(x)\phi'(x)-f'(x)\phi(x) \right \}}{f(x)\phi(x)}\left \{ \log\phi(x)-\log f(x) \right \}dx$

$\displaystyle I=\int \frac { \left\{ f(x)\phi '(x)-f'(x)\phi (x) \right\} }{ f(x)\phi (x) } \left\{ \log\frac { \phi (x) }{ f(x) } \right\} dx$

$\displaystyle =\int \log\frac{\phi(x)}{f(x)}d\left \{ \log\frac{\phi(x)}{f(x)} \right \}dx$

$\displaystyle =\frac{1}{2}\left \{ \log\frac{\phi(x)}{f(x)} \right \}^{2}+c$
Hence, statement 1 is true.

Consider, $I= \int (h(x))^{ n }h'(x)dx$
Put $h(x)=t$
$\Rightarrow \displaystyle h'(x)dx=dt$

$\therefore I=\int t^{n} dt$
$=\displaystyle \frac{t^{n+1}}{n+1}+C$

So, $I=\displaystyle \frac { (h(x))^{ n+1 } }{ n+1 } +c$
Hence, statement 2 is correct and is the correct explanation of statement 1

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Assertion & Reason Hard Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 105

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