To solve the integral

$\int_0^1 \frac{18x^{17}}{x^{18} + 3} \, dx,$

we proceed as follows:

Step 1: Simplify the expression

The integrand is:

$\frac{18x^{17}}{x^{18} + 3}.$

Step 2: Substitution

Let:

$u = x^{18} + 3, \quad \text{so that } \frac{du}{dx} = 18x^{17}.$

Thus:

$du = 18x^{17} \, dx.$

When $x = 0$, $u = 3$.

When $x = 1$, $u = 1^{18} + 3 = 4$.

Step 3: Update the integral

Substituting $u$ and $du$, the integral becomes:

$\int_3^4 \frac{1}{u} \, du.$

Step 4: Evaluate the integral

The integral of $\frac{1}{u}$ is:

$\int \frac{1}{u} \, du = \ln|u| + C.$

Thus:

$\int_3^4 \frac{1}{u} \, du = \ln(4) - \ln(3) = \ln\left(\frac{4}{3}\right).$

Final Answer:

$\int_0^1 \frac{18x^{17}}{x^{18} + 3} \, dx = \ln\left(\frac{4}{3}\right).$

Would you like a more detailed explanation, or do you have any questions?

Here are 5 related questions for practice:

1. Evaluate $\int_0^1 \frac{x^4}{x^5 + 2} \, dx$ using substitution.
2. Solve $\int_0^1 \frac{1}{x^2 + 1} \, dx$.
3. Find the integral of $\int_a^b \frac{1}{u} \, du$.
4. Evaluate $\int_0^1 x^{17} \, dx$.
5. Solve $\int_1^e \ln(x) \, dx$.

Tip: When tackling integrals involving rational functions, substitution often simplifies the problem significantly.