Question

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

• A. 32
• B. 16
• C. 8
• D. 4

Answer 1

Answer: (A)

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