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

Ask by Donnelly Donnelly. in South Africa

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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