Indefinite Integration
Sixth Root Substitution — Polynomial Division
nta_pyq_2026_jan
Grade 12

Question:

Let $f(x)=\displaystyle\int\frac{dx}{x^{2/3}+2x^{1/2}}$ be such that $f(0)=-26+24\log_e 2$. If $f(1)=a+b\log_e 3$, where $a,b\in\mathbb{Z}$, then $a+b$ is equal to:
-26
-11
-5
-18

Step-by-Step Solution

Key Concept: Let $u=x^{1/6}$, so $x=u^6$, $dx=6u^5du$. Then $x^{2/3}=u^4$, $x^{1/2}=u^3$. Integral $=\int\tfrac{6u^5du}{u^4+2u^3}=\int\tfrac{6u^2du}{u+2}$. Polynomial division: $\tfrac{u^2}{u+2}=u-2+\tfrac{4}{u+2}$.
$f(1)=-35+24\ln3$. $a+b=-11$.
Correct Answer: 2

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