JEE Main 2026 — Mathematics Calculus

Let $u = x^{1/6}$u=x1/6, so $x = u^6$x=u6 and $dx = 6u^5 du$dx=6u5du. Then $x^{2/3} = u^4$x2/3=u4, $x^{1/2} = u^3$x1/2=u3.

$\displaystyle f(x) = \int \frac{6u^5 du}{u^4 + 2u^3} = \int \frac{6u^2 du}{u+2}$f(x)=∫u4+2u36u5du​=∫u+26u2du​.

By polynomial division: $\frac{u^2}{u+2} = u - 2 + \frac{4}{u+2}$u+2u2​=u−2+u+24​.

$f(x) = 6\left(\frac{u^2}{2} - 2u + 4\ln|u+2|\right) + C = 3u^2 - 12u + 24\ln(u+2) + C$f(x)=6(2u2​−2u+4ln∣u+2∣)+C=3u2−12u+24ln(u+2)+C.

Substituting back: $f(x) = 3x^{1/3} - 12x^{1/6} + 24\ln(x^{1/6}+2) + C$f(x)=3x1/3−12x1/6+24ln(x1/6+2)+C.

From $f(0) = -26 + 24\ln(2)$f(0)=−26+24ln(2): $24\ln(2) + C = -26 + 24\ln(2)$24ln(2)+C=−26+24ln(2), so $C = -26$C=−26.

At $x = 1$x=1: $f(1) = 3 - 12 + 24\ln(3) - 26 = -35 + 24\ln(3)$f(1)=3−12+24ln(3)−26=−35+24ln(3).

Thus $a = -35$a=−35, $b = 24$b=24, and $a + b = -11$a+b=−11.