Integral Calculus – Indefinite Integration – JEE Main 28 Jan 2025 Shift 2

Solution:

Let $I = \int\frac{1}{x^{1/4}(1+x^{1/4})}dx$.

Put $x = t^{4} \Rightarrow dx = 4t^{3}dt$.

Substituting in the integral:
$$I = \int \frac{4t^{3}}{t(1+t)} dt = \int \frac{4t^{2}}{1+t} dt$$

$$I = 4 \int \frac{t^{2}-1+1}{1+t} dt$$

$$I = 4 \int \left( \frac{(t-1)(t+1)}{1+t} + \frac{1}{1+t} \right) dt$$

$$I = 4 \int \left( t-1 + \frac{1}{1+t} \right) dt$$

$$I = 4 \left( \frac{t^{2}}{2} – t + \ln|t+1| \right) + C$$

Substitute $t = x^{1/4}$:
$$f(x) = 2(x^{1/4})^2 – 4x^{1/4} + 4\ln(x^{1/4}+1) + C$$
$$f(x) = 2x^{1/2} – 4x^{1/4} + 4\ln(1+x^{1/4}) + C$$

Given $f(0) = -6$:
$$f(0) = 0 – 0 + 4\ln(1) + C = -6 \Rightarrow C = -6$$

Now find $f(1)$:
$$f(1) = 2(1) – 4(1) + 4\ln(1+1) – 6$$
$$f(1) = 2 – 4 + 4\ln 2 – 6$$
$$f(1) = 4\ln 2 – 8$$
$$f(1) = 4(\ln 2 – 2)$$

Ans. (1)