Integral Calculus – Indefinite Integration – JEE Main 28 Jan 2026 Shift 2

Solution:

$$f(x) = \int \frac{dx}{x^{2/3} + 2x^{1/2}}$$

To eliminate the fractional powers, let $x = t^6$ (since 6 is the LCM of 3 and 2).
$$dx = 6t^5 dt$$

Substitute these into the integral:
$$f(x) = \int \frac{6t^5 dt}{(t^6)^{2/3} + 2(t^6)^{1/2}}$$
$$f(x) = \int \frac{6t^5 dt}{t^4 + 2t^3}$$

Factor out $t^3$ from the denominator:
$$f(x) = \int \frac{6t^5 dt}{t^3(t + 2)} = 6 \int \frac{t^2}{t + 2} dt$$

Add and subtract 4 in the numerator to simplify:
$$f(x) = 6 \int \frac{t^2 – 4 + 4}{t + 2} dt$$
$$f(x) = 6 \int \left( \frac{(t-2)(t+2)}{t+2} + \frac{4}{t+2} \right) dt$$
$$f(x) = 6 \int \left( t – 2 + \frac{4}{t+2} \right) dt$$

Integrate each term:
$$f(x) = 6 \left[ \frac{t^2}{2} – 2t + 4 \ln|t+2| \right] + C$$
$$f(x) = 3t^2 – 12t + 24 \ln|t+2| + C$$

Substitute back $t = x^{1/6}$:
$$f(x) = 3x^{1/3} – 12x^{1/6} + 24 \ln|x^{1/6} + 2| + C$$

Use the given condition $f(0) = -26 + 24\log_e(2)$:
$$f(0) = 3(0) – 12(0) + 24 \ln(0 + 2) + C$$
$$24 \ln 2 + C = -26 + 24 \ln 2$$
$$C = -26$$

The function is $f(x) = 3x^{1/3} – 12x^{1/6} + 24 \ln|x^{1/6} + 2| – 26$.

Now evaluate $f(1)$:
$$f(1) = 3(1)^{1/3} – 12(1)^{1/6} + 24 \ln(1^{1/6} + 2) – 26$$
$$f(1) = 3 – 12 + 24 \ln(3) – 26$$
$$f(1) = -35 + 24 \ln 3$$

Compare with $f(1) = a + b\log_e(3)$:
$$a = -35, \quad b = 24$$

Calculate $a+b$:
$$a + b = -35 + 24 = -11$$

Ans. (3)