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: Approach
Check that the derivative of option (D) gives the integrand.

Step 2: Differentiate
Let $u=2x^{21}+7x^6+14x^3$ and $F=\dfrac{u^{4/3}}{56}$. Then $F'=\dfrac{1}{56}\cdot\dfrac43u^{1/3}u'=\dfrac{u^{1/3}u'}{42}$.

Step 3: Substitute u'
$u'=42x^2(x^{18}+x^3+1)$, so $F'=x^2(x^{18}+x^3+1)u^{1/3}$.

Step 4: Match
Since $u^{1/3}=x(2x^{18}+7x^3+14)^{1/3}$, we get $F'=(x^{21}+x^6+x^3)(2x^{18}+7x^3+14)^{1/3}$, the integrand. So (D).

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