Question

If $\int (x^{24}+x^{16}+x^8)(2x^{16}+3x^8+6)^{\frac{1}{8}}dx =$

$\frac{1}{\alpha}(2x^{24}+3x^{16}+6x^8)^{\frac{\beta}{\gamma}}+C$ (where C is constant of integration and $\beta, \gamma$ are coprime numbers), then the value of $(\alpha+\beta+\gamma)$ is:

• A. 61
• B. 67
• C. 71
• D. 59

Answer 1

Answer: (C)

71

Answer 2

To evaluate the integral $\int (x^{24}+x^{16}+x^8)(2x^{16}+3x^8+6)^{\frac{1}{8}}dx$, we use a substitution method.

Let $u = 2x^{24}+3x^{16}+6x^8$. Then $\frac{du}{dx} = 48x^{23} + 48x^{15} + 48x^7 = 48(x^{23} + x^{15} + x^7)$ So, $(x^{23} + x^{15} + x^7)dx = \frac{1}{48}du$.

The integral becomes: $I = \int (x^{23}+x^{15}+x^7) (2x^{24}+3x^{16}+6x^8)^{\frac{1}{8}} dx = \int u^{\frac{1}{8}} \left(\frac{1}{48}du\right) = \frac{1}{48} \int u^{\frac{1}{8}} du$ $I = \frac{1}{48} \left( \frac{u^{\frac{1}{8}+1}}{\frac{1}{8}+1} \right) + C = \frac{1}{48} \left( \frac{u^{\frac{9}{8}}}{\frac{9}{8}} \right) + C = \frac{1}{48} \cdot \frac{8}{9} u^{\frac{9}{8}} + C = \frac{1}{54} u^{\frac{9}{8}} + C$

Substituting back $u = 2x^{24}+3x^{16}+6x^8$: $I = \frac{1}{54} (2x^{24}+3x^{16}+6x^8)^{\frac{9}{8}} + C$

Comparing with the given form $\frac{1}{\alpha}(2x^{24}+3x^{16}+6x^8)^{\frac{\beta}{\gamma}}+C$, we have: $\alpha = 54$ $\frac{\beta}{\gamma} = \frac{9}{8}$

Since $\beta$ and $\gamma$ are coprime, $\beta = 9$ and $\gamma = 8$.

Therefore, $\alpha+\beta+\gamma = 54 + 9 + 8 = 71$.