Sure, let's evaluate the integral [tex]\(\int x^2 12^{x^3+3} \, dx\)[/tex].
To solve this, observe the integral and consider if we can use a substitution method. A useful substitution is often connected to the inner function, in this case, [tex]\(x^3 + 3\)[/tex].
Let:
[tex]\[ u = x^3 + 3 \][/tex]
Then, differentiate [tex]\(u\)[/tex] with respect to [tex]\(x\)[/tex]:
[tex]\[ \frac{du}{dx} = 3x^2 \][/tex]
[tex]\[ du = 3x^2 \, dx \][/tex]
[tex]\[ \frac{1}{3} du = x^2 \, dx \][/tex]
The integral becomes:
[tex]\[ \int x^2 12^{x^3+3} \, dx = \int 12^u \cdot \frac{1}{3} \, du \][/tex]
[tex]\[ = \frac{1}{3} \int 12^u \, du \][/tex]
Next, we integrate [tex]\(12^u\)[/tex] with respect to [tex]\(u\)[/tex]. Recall that the integral of [tex]\(a^u \, du\)[/tex] is [tex]\(\frac{a^u}{\ln(a)}\)[/tex], where [tex]\(a \neq 1\)[/tex].
Thus:
[tex]\[ \int 12^u \, du = \frac{12^u}{\ln(12)} \][/tex]
Substituting back [tex]\(u = x^3 + 3\)[/tex], we get:
[tex]\[ \int 12^u \cdot \frac{1}{3} \, du = \frac{1}{3} \cdot \frac{12^u}{\ln(12)} \][/tex]
Finally, substituting [tex]\(u = x^3 + 3\)[/tex] back in, we obtain:
[tex]\[ \frac{1}{3} \cdot \frac{12^{x^3 + 3}}{\ln(12)} = \frac{12^{x^3 + 3}}{3 \ln(12)} \][/tex]
So, the evaluated integral is:
[tex]\[ \int x^2 12^{x^3 + 3} \, dx = \frac{12^{x^3 + 3}}{3 \ln(12)} \][/tex]
This is the final solution:
[tex]\[ \int x^2 12^{x^3+3} \, dx = \boxed{\frac{12^{x^3 + 3}}{3 \ln(12)}} \][/tex]
