Question

\(\int (x^{21}+x^6+x^3)(2x^{18}+7x^3+14)^{\frac{1}{3}} dx =\)

• A. \(\frac{1}{56}(2x^{18}+7x^3+14)^{\frac{4}{3}}+c\)
• B. \((2x^{18}+7x^3+14)^{\frac{4}{3}}+c\)
• C. \((2x^{21}+7x^6+14x^3)^{\frac{4}{3}}+c\)
• D. \(\frac{1}{56}(2x^{21}+7x^6+14x^3)^{\frac{4}{3}}+c\)

Hint:

Factor x cubed from the bracket to spot that the integrand is a constant times u to the one third times du.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept
Try the substitution \(u=2x^{21}+7x^6+14x^3\), because its derivative looks like the bracket in the integrand.

Step 2: Differentiate u
\[ du=(42x^{20}+42x^5+42x^2)\,dx=42x^2(x^{18}+x^3+1)\,dx \]

Step 3: Rewrite the integrand
\[ x^{21}+x^6+x^3=x^3(x^{18}+x^3+1) \]
\[ 2x^{18}+7x^3+14=\frac{u}{x^3} \]
So the integrand is
\[ x^3(x^{18}+x^3+1)\cdot\frac{u^{1/3}}{x}=x^2(x^{18}+x^3+1)\,u^{1/3} \]

Step 4: Integrate
\[ \int x^2(x^{18}+x^3+1)u^{1/3}dx=\frac1{42}\int u^{1/3}du=\frac1{42}\cdot\frac34u^{4/3}=\frac{u^{4/3}}{56} \]
The result is option (D).

Final Answer:
With u = 2x^21 + 7x^6 + 14x^3 the integral is u^(4/3)/56, option (D). \[ \boxed{\frac{1}{56}(2x^{21}+7x^6+14x^3)^{4/3}+c} \]