Question

$$int 18x^3(3x^4 + 7)^9 dx$$
a.) $\frac{9}{5}(3x^4 + 7)^{10}+c$
b.) $\frac{3}{20}u^{10}+c$
c.) $\frac{9}{20}x^4(3x^4 + 7)^{10}+c$
d.) $\frac{3}{20}(3x^4 + 7)^{10}+c$

Explanation:

Step1: Choose substitution variable

Let $u = 3x^4 + 7$

Step2: Compute derivative of u

$\frac{du}{dx} = 12x^3 \implies du = 12x^3 dx \implies x^3 dx = \frac{du}{12}$

Step3: Rewrite integral in terms of u

$\int 18x^3(3x^4+7)^9 dx = 18 \int u^9 \cdot \frac{du}{12} = \frac{18}{12} \int u^9 du = \frac{3}{2} \int u^9 du$

Step4: Integrate using power rule

$\frac{3}{2} \cdot \frac{u^{10}}{10} + C = \frac{3}{20}u^{10} + C$

Step5: Substitute back u

$\frac{3}{20}(3x^4 + 7)^{10} + C$

Answer:

d.) $\frac{3}{20}(3x^4 + 7)^{10}+c$