Let f ( x ) = 2 x ( x 2 + 1 ) ( x 2 + 3 ) d x . If f ( 3 ) = 1 2 ( log e 5 - log e 6 ) , then f ( 4 ) is equal to [2023]

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Q.
$Let f (x) = \int \frac{2 x}{(x^{2} + 1) (x^{2} + 3)} d x .$ $If f (3) = \frac{1}{2} (\log_{e} 5 - \log_{e} 6), then f (4) is equal to$ [2023]

1 $\log_{e} 17 - \log_{e} 18$

2 $\log_{e} 19 - \log_{e} 20$

3 $\frac{1}{2} (\log_{e} 19 - \log_{e} 17)$

4 $\frac{1}{2} (\log_{e} 17 - \log_{e} 19)$

Ans.
(4)

$Let x^{2} = t, 2 x d x = d x$

$f (x) = \int \frac{d t}{(t + 1) (t + 3)} = \frac{1}{2} \int (\frac{1}{t + 1} - \frac{1}{t + 3}) d t$

$f (x) = \frac{1}{2} \log_{e} (\frac{t + 1}{t + 3}) + c = \frac{1}{2} \log_{e} (\frac{x^{2} + 1}{x^{2} + 3}) + c$

$f (3) = \frac{1}{2} \log_{e} (\frac{10}{12}) + c = \frac{1}{2} \log_{e} \frac{5}{6} + c = \frac{1}{2} (\log_{e} 5 - \log_{e} 6) \Rightarrow c = 0$

$f (x) = \frac{1}{2} \log_{e} (\frac{x^{2} + 1}{x^{2} + 3}); f (4) = \frac{1}{2} \log_{e} (\frac{17}{19}) = \frac{1}{2} (\log_{e} 17 - \log_{e} 19)$

Solution

Q.
$Let f (x) = \int \frac{2 x}{(x^{2} + 1) (x^{2} + 3)} d x .$ $If f (3) = \frac{1}{2} (\log_{e} 5 - \log_{e} 6), then f (4) is equal to$ [2023]

1 $\log_{e} 17 - \log_{e} 18$

2 $\log_{e} 19 - \log_{e} 20$

3 $\frac{1}{2} (\log_{e} 19 - \log_{e} 17)$

4 $\frac{1}{2} (\log_{e} 17 - \log_{e} 19)$

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Solution

Q. Let f(x)=∫2x(x2+1)(x2+3)dx. If f(3)=12(loge5-loge6), then f(4) is equal to [2023]

1 loge17-loge18

2 loge19-loge20

3 12(loge19-loge17)

4 12(loge17-loge19)

Ans. (4) Let x2=t, 2x dx=dx f(x)=∫dt(t+1)(t+3)=12∫(1t+1-1t+3)dt f(x)=12loge(t+1t+3)+c=12loge(x2+1x2+3)+c f(3)=12loge(1012)+c =12loge56+c=12(loge5-loge6)⇒c=0 f(x)=12loge(x2+1x2+3); f(4)=12loge(1719)=12(loge17-loge19)

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Q.
$Let f (x) = \int \frac{2 x}{(x^{2} + 1) (x^{2} + 3)} d x .$ $If f (3) = \frac{1}{2} (\log_{e} 5 - \log_{e} 6), then f (4) is equal to$ [2023]

1 $\log_{e} 17 - \log_{e} 18$

2 $\log_{e} 19 - \log_{e} 20$

3 $\frac{1}{2} (\log_{e} 19 - \log_{e} 17)$

4 $\frac{1}{2} (\log_{e} 17 - \log_{e} 19)$