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`int ((f (x) g' (x) -f' (x) g (x))/(f (x) g(x)))(log (g(x )) - log (f(x)))dx `is equal to:

A

`log_(e){(g(x))/(f(x))}+C`

B

`(1)/(2){"log"_(e)(g(x))/(f(x))}^(2)+C`

C

`(g(x))/(f(x))"log"_(e)(g(x))/(f(x))+C`

D

none of these

Text Solution

Verified by Experts

The correct Answer is:
b

Let `I=int(f(x)g'(x)-g(x)f'(x))/(f(x)*g(x)){log(x)-logf(x)}dx`
`rArrI=intlog{(g(x))/(f(x))}*d{"log"(g(x))/(f(x))}=(1)/(2){"log"_(e)(g(x))/(f(x))}^(2)+C`
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OBJECTIVE RD SHARMA ENGLISH-INDEFINITE INTEGRALS-Solved Example
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