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JEE Main 202528 JanuaryEvening Shift
Exam
JEE Main
Year
2025
Date
28 January
Shift
Evening Shift
Subject
Mathematics
Chapter
Indefinite Integration

JEE Main 2025 · 28 January · Evening Shift · Mathematics · Indefinite Integration

If f(x)=∫ 1 x^ 1 / 4 (1+x^ 1 / 4 ) d x, f(0)=-6 , then f(1) is...

If f(x)=∫1x1/4(1+x1/4)dx,f(0)=−6f(x)=\int \frac{1}{x^{1 / 4}\left(1+x^{1 / 4}\right)} \mathrm{d} x, f(0)=-6, then f(1)f(1) is equal to :

Options

  1. A4(log⁡e2−2)4\left(\log _e 2-2\right)
  2. B2−log⁡e222-\log _{e^2} 2
  3. Clog⁡e2+2\log _{\mathrm{e}} 2+2
  4. D4(log⁡e2+2)4\left(\log _e 2+2\right)

Correct answer

A. 4(log⁡e2−2)4\left(\log _e 2-2\right)

Step-by-step solution

 Put x1/4=t⇒dx=4t3dt∫4t3dtt(t+1)=4∫(t2−1t+1+1t+1)dtf(x)=4[x1/22−x1/4+ln⁡∣x1/4+1∣]+C\begin{aligned} & \text{ 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\left(\frac{t^2-1}{t+1}+\frac{1}{t+1}\right) d t \\ & f(x)=4\left[\frac{x^{1 / 2}}{2}-x^{1 / 4}+\ln \left|x^{1 / 4}+1\right|\right]+C\end{aligned}f(0)=−6⇒C=−6⇒f(1)=4(log⁡e2−2)\begin{aligned} & f(0)=-6 \\ & \Rightarrow C=-6 \\ & \Rightarrow \quad f(1)=4\left(\log _e 2-2\right)\end{aligned}$
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