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
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