Question

If $\int\left(x^9+x^6+x^3\right)\left(2 x^6+3 x^3+6\right)^{\frac{1}{3}} d x=\frac{1}{A}\left(2 x^9+3 x^6+6 x^3\right)^B+C$, where $C$ is integration constant then $AB$ is equal to

• A. 32
• B. 16
• C. 8
• D. 4

Answer 1

Answer: (A)

$\Theta \int\left(x^9+x^6+x^3\right)\left(2 x^6+3 x^3+6\right)^{\frac{1}{3}} d x $
$=\int\left(x^8+x^5+x^2\right)\left(2 x^9+3 x^6+6 x^3\right)^{\frac{1}{3}} d x$
Let $ 2 x^9+3 x^6+6 x^3=t$
$\Rightarrow 18\left( x ^8+ x ^5+ x ^2\right) dx = dt$
$\therefore I =\int \frac{ t ^{1 / 3}}{18} dt =\frac{1}{18} \cdot \frac{ t ^{4 / 3}}{4 / 3}+ C =\frac{1}{24} t ^{4 / 3}+ C$
$\therefore AB =24 \times \frac{4}{3}=32$