QJEEM25Ma7f56: JEE Main 2025 Mathematics - Class 12 JEEM25Ma7f56 Solution: If , then is equal to :

Substitution: Start by substituting $x = t^4$ which implies $dx = 4t^3 \, dt$ .

Integral Transformation:

$\int \frac{1}{x^{1/4}(1 + x^{1/4})} \, dx \Rightarrow \int \frac{4t^3 \, dt}{t(1 + t)} = \int \frac{4t}{1 + t} \, dt$

Rewriting the Integral: The expression $\frac{4t}{1 + t}$ can be decomposed as:

$\frac{4t}{1 + t} = 4\left(\frac{t^2 - 1 + 1}{1 + t}\right) = 4\left((t - 1) + \frac{1}{t + 1}\right)$

Separate and Integrate:

$4 \int (t - 1) \, dt + 4 \int \frac{1}{t + 1} \, dt$

Solving the Integrals:

The integral of $t - 1$ is $\frac{(t - 1)^2}{2}$ .

The integral of $\frac{1}{t + 1}$ is $\ln(t + 1)$ .

Therefore:

$f(x) = 4 \left\{ \frac{(t - 1)^2}{2} + \ln(t + 1) \right\} + C$

Substitute $t = x^{1/4}$ : Replace $t$ back with $x^{1/4}$ :

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

Using the Condition $f(0) = -6$ :

$f(0) = 2(0 - 1)^2 + 4 \ln(1 + 0) + C = 2 \times 1 - 0 + C = -6$

Solving gives $C = -8$ .

Find $f(1)$ :

$f(1) = 2(1^{1/4} - 1)^2 + 4 \ln(1 + 1^{1/4}) - 8$

Simplifies to:

$f(1) = 0 + 4 \ln 2 - 8 = 4(\ln 2 - 2)$