If f ( x ) = ∫ x 1/4 ( 1 + x 1/4 ) 1 d x , f ( 0 ) = − 6 , then f ( 1 ) is equal to :

Put x 1/4 = t ⇒ d x = 4 t 3 d t ∫ t ( t + 1 ) 4 t 3 d t = 4 ∫ ( t + 1 t 2 − 1 + t + 1 1 ) d t f ( x ) = 4 [ 2 x 1/2 − x 1/4 + ln x 1/4 + 1 ] + C f ( 0 ) = − 6 ⇒ C = − 6 ⇒ f ( 1 ) = 4 ( lo g e 2 − 2 )

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Mathematics - Indefinite Integration Question with Solution | TestHub

MathematicsIndefinite IntegrationGeneralMedium2 minPYQ_2025

MathematicsMediumsingle choice

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

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Solution:

$Put x^{1/4} = t \Rightarrow d x = 4 t^{3} d t \int \frac{4 t ^{3} d t}{t ( t + 1 )} = 4 \int (\frac{t ^{2} - 1}{t + 1} + \frac{1}{t + 1}) d t f (x) = 4 [\frac{x ^{1/2}}{2} - x^{1/4} + ln x^{1/4} + 1] + C$
$f (0) = - 6 \Rightarrow C = - 6 \Rightarrow f (1) = 4 (lo g_{e} 2 - 2)$

Stream:JEESubject:MathematicsTopic:Indefinite IntegrationSubtopic:General

⏱ 2mℹ️ Source: PYQ_2025

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