If , f(0) = – 6, then f(1) is equal to : [2025]

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Solution

Q.
If $f (x) = \int \frac{1}{x^{1 / 4} (1 + x^{1 / 4})} d x$ , f(0) = – 6, then f(1) is equal to : [2025]

1 $\log_{e} 2 + 2$

2 $4 (\log_{e} 2 - 2)$

3 $4 (\log_{e} 2 + 2)$

4 $2 - \log_{e} 2$

Ans.
(2)

We have, $f (x) = \int \frac{1}{x^{1 / 4} (1 + x^{1 / 4})} d x$

Putting $x = t^{4} \Rightarrow d x = 4 t^{3} d t$

$f (t^{4}) = \int \frac{4 t^{3}}{t (1 + t)} d t = 4 \int \frac{t^{2}}{1 + t} d t$

$= 4 \int \frac{(t^{2} - 1) + 1}{1 + t} d t$

$= 4 [\int (t - 1) d t + \int \frac{1}{t + 1} d t]$

$= 4 [\frac{t^{2}}{2} - t + \log_{e} (t + 1)] + C$

$∴ f (x) = 2 x^{1 / 2} - 4 x^{1 / 4} + 4 \log_{e} (x^{1 / 4} + 1) + C$

$f (0) = - 6 \Rightarrow C = - 6$

$∴ f (1) = 2 - 4 + 4 \log_{e} 2 - 6 = 4 \log_{e} 2 - 8 = 4 (\log_{e} 2 - 2)$

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