Indefinite Integration
Substitution in complex integrand
MJAT_TS1_P1
Grade 12
Question:
$$\int_1 \frac{(2x^3 + 3x^2 - 1)\,\sqrt{4x^6 + 12x^5 + 9x^4 - 6x^3 - 9x^2 + 2}}{x(x+1)}\,dx =$$
A) $\dfrac{2\sqrt{78} + 9\sqrt{3}}{54}$
B) $\dfrac{2\sqrt{78} - 9\sqrt{3}}{54}$
C) $\dfrac{9\sqrt{3} - 2\sqrt{78}}{54}$
D) $\dfrac{2\sqrt{26} - 3\sqrt{3}}{18}$
Step-by-Step Solution
Key Concept: Let $t = 2x^3 + 3x^2 - 1$. The expression under the square root factors to $(2x^3+3x^2-1)^2 - (2x^3+3x^2-1) + 1$ (approximately). The substitution reduces to $\int \frac{\sqrt{t^2 + \cdots}}{\cdots}\,dt$.
Let $t = 2x^3 + 3x^2 - 1$, $dt = (6x^2+6x)dx = 6x(x+1)dx$. The radicand becomes $t^2 - t + 1$ (after simplification). Integral reduces to $\frac{1}{6}\int \frac{\sqrt{t^2-t+1}}{1}\,dt$, evaluated at bounds to give $\dfrac{2\sqrt{78}-9\sqrt{3}}{54}$.
Correct Answer: B