Evaluate the following integrals using the appropriate rules of integration. Show all your calculations in each cas
a.

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Mar 28,2025

Question

Evaluate the following integrals using the appropriate rules of integration. Show all your calculations in each cas a. $$ \int \frac{t^{2}+8}{t^{2}-5 t+6} d t $$

UpStudy AI · Accepted Answer

We begin by evaluating

$$
\int \frac{t^2+8}{t^2-5t+6}\,dt.
$$

**Step 1. Factor the denominator.**

The quadratic in the denominator factors as

$$
t^2-5t+6=(t-2)(t-3).
$$

Thus, the integral becomes

$$
\int \frac{t^2+8}{(t-2)(t-3)}\,dt.
$$

**Step 2. Perform polynomial long division (if needed).**

Since the degree of the numerator $$t^2+8$$ is equal to the degree of the denominator $$t^2-5t+6$$, perform the division:

Dividing $$t^2+8$$ by $$t^2-5t+6$$:

- The quotient is $$1$$ because $$t^2 / t^2 = 1$$.
- The remainder $$R(t)$$ is given by

$$
R(t)=\bigl(t^2+8\bigr)-\bigl(t^2-5t+6\bigr)=5t+2.
$$

Thus,

$$
\frac{t^2+8}{t^2-5t+6}=1+\frac{5t+2}{(t-2)(t-3)}.
$$

So the integral becomes

$$
\int \frac{t^2+8}{t^2-5t+6}\,dt=\int 1\,dt+\int \frac{5t+2}{(t-2)(t-3)}\,dt.
$$

**Step 3. Decompose the proper rational function into partial fractions.**

We write

$$
\frac{5t+2}{(t-2)(t-3)}=\frac{A}{t-2}+\frac{B}{t-3}.
$$

Multiply both sides by $$(t-2)(t-3)$$:

$$
5t+2=A(t-3)+B(t-2).
$$

Expanding the right-hand side:

$$
5t+2=At-3A+Bt-2B=(A+B)t-(3A+2B).
$$

Now equate the coefficients of like terms:

1. Coefficient of $$t$$: 
   $$
   A+B=5.
   $$
2. Constant term: 
   $$
   -(3A+2B)=2 \quad \Longrightarrow \quad 3A+2B=-2.
   $$

Solve for $$A$$ and $$B$$:

From $$A+B=5$$, we have $$A=5-B$$. Substitute into $$3A+2B=-2$$:

$$
3(5-B)+2B=-2.
$$

Simplify:

$$
15-3B+2B=-2 \quad \Longrightarrow \quad 15-B=-2.
$$

Thus,

$$
-B=-17 \quad\Longrightarrow\quad B=17.
$$

Then,

$$
A=5-B=5-17=-12.
$$

So the partial fraction decomposition is

$$
\frac{5t+2}{(t-2)(t-3)}=-\frac{12}{t-2}+\frac{17}{t-3}.
$$

**Step 4. Integrate term by term.**

Returning to the integral:

$$
\int \frac{t^2+8}{t^2-5t+6}\,dt = \int 1\,dt -12\int \frac{1}{t-2}\,dt+17\int \frac{1}{t-3}\,dt.
$$

Integrate each term:

1. $$\int 1\,dt=t.$$
2. $$\int \frac{1}{t-2}\,dt=\ln |t-2|.$$
3. $$\int \frac{1}{t-3}\,dt=\ln |t-3|.$$

Thus, the result is

$$
t-12\ln|t-2|+17\ln|t-3|+C.
$$

**Final Answer**

$$
\boxed{t-12\ln|t-2|+17\ln|t-3|+C}
$$
$$ \int \frac{t^2+8}{t^2-5t+6}\,dt = t - 12\ln|t-2| + 17\ln|t-3| + C $$

Evaluate the following integrals using the appropriate rules of integration. Show all your calculations in each cas
a.

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