Question

1. \\( \int \frac{1}{\sqrt{1 - t^2}} dt \\)\
2. \\( \int \frac{x^3 - 3x^2 + 5}{x - 3} dx \\)

Explanation:

Problem 5:

Step1: Recall Integral Formula

The integral of $\frac{1}{\sqrt{1 - u^2}}$ with respect to $u$ is a standard integral. The formula is $\int \frac{1}{\sqrt{1 - u^2}} du=\arcsin(u)+C$, where $C$ is the constant of integration.

Step2: Apply the Formula

In the given integral $\int \frac{1}{\sqrt{1 - t^2}} dt$, we can directly apply the above formula by substituting $u = t$. So the integral becomes $\arcsin(t)+C$.

Step1: Polynomial Long Division

We divide the numerator $x^3 - 3x^2+5$ by the denominator $x - 3$.
Using polynomial long division:

• Divide $x^3$ by $x$ to get $x^2$. Multiply $(x - 3)$ by $x^2$ to get $x^3-3x^2$.
• Subtract $x^3 - 3x^2$ from $x^3 - 3x^2+5$: $(x^3 - 3x^2+5)-(x^3 - 3x^2)=5$.

So, $\frac{x^3 - 3x^2+5}{x - 3}=x^2+\frac{5}{x - 3}$.

Step2: Integrate Term by Term

Now we integrate $x^2+\frac{5}{x - 3}$ with respect to $x$.

• The integral of $x^2$ with respect to $x$ is $\frac{x^{2 + 1}}{2+1}=\frac{x^3}{3}$.
• The integral of $\frac{5}{x - 3}$ with respect to $x$: using the substitution $u=x - 3$, $du=dx$, so $\int\frac{5}{x - 3}dx = 5\int\frac{1}{u}du=5\ln|u|+C = 5\ln|x - 3|+C$.
• Combining these results, the integral is $\frac{x^3}{3}+5\ln|x - 3|+C$.

Answer:

$\arcsin(t)+C$ (where $C$ is the constant of integration)

Problem 7: