JEE Main 2025 Jan 28 Shift 2, Mathematics Q11: Methods of Integration

JEE Main2025Jan 28, Shift 2Mathematics

If $f (x) = \int \frac{1}{x ^{1/4} ( 1 + x ^{1/4} )} d x$ , $f (0) = - 6$ , then $f (1)$ is equal to :

A
$lo g_{e} 2 + 2$
B
$4 (lo g_{e} 2 - 2)$
C
$2 - lo g_{e} 2$
D
$4 (lo g_{e} 2 + 2)$

Solution

Substitute $x = t^{4}$ , $d x = 4 t^{3} d t$ :

$\int \frac{4 t ^{3} d t}{t ( 1 + t )} = \int \frac{4 t ^{2}}{1 + t} d t = 4 \int (t - 1 + \frac{1}{1 + t}) d t = 4 (\frac{( t - 1 ) ^{2}}{2} + ln (1 + t)) + C^{'}$ (after grouping).

Back-substituting $t = x^{1/4}$ :

$f (x) = 2 (x^{1/4} - 1)^{2} + 4 ln (1 + x^{1/4}) + C$ .

Using $f (0) = - 6$ : $2 (1) + 0 + C = - 6 \Rightarrow C = - 8$ .

At $x = 1$ : $f (1) = 2 (0) + 4 ln 2 - 8 = 4 ln 2 - 8 = 4 (ln 2 - 2)$ .

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JEE Main 2025 Jan 28 Shift 2, Mathematics Q11: Methods of Integration

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