$$ \begin{array}{ll} ext { 3.1 Sparks and dust generated by grinding. } \ 3.1 & \int \frac{x^{2}+1}{x^{3}+x^{2}} d x\end{array} $$

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$$ \begin{array}{ll}	ext { 3.1 Sparks and dust generated by grinding. } \ 3.1 & \int \frac{x^{2}+1}{x^{3}+x^{2}} d x\end{array} $$

UpStudy AI · Accepted Answer

We start with the integral

$$
\int \frac{x^2+1}{x^3+x^2}\,dx.
$$

**Step 1. Factor the denominator**

Notice that

$$
x^3+x^2=x^2(x+1).
$$

Thus, the integrand becomes

$$
\frac{x^2+1}{x^2(x+1)}.
$$

**Step 2. Partial Fractions Decomposition**

We set

$$
\frac{x^2+1}{x^2(x+1)}=\frac{A}{x}+\frac{B}{x^2}+\frac{C}{x+1}.
$$

Multiply both sides by $$x^2(x+1)$$ to clear the denominators:

$$
x^2+1 = A\,x(x+1) + B(x+1) + Cx^2.
$$

Expand the right side:

$$
x^2+1 = A x^2 + A x + B x + B + C x^2.
$$

Group like terms:

$$
x^2+1 = (A+C)x^2 + (A+B)x + B.
$$

Now, equate the coefficients from both sides:

- Coefficient of $$x^2$$: $$A+C = 1$$.
- Coefficient of $$x$$: $$A+B = 0$$.
- Constant term: $$B = 1$$.

From $$B=1$$, we have $$A = -B = -1$$. Then, using $$A+C=1$$:

$$
-1+C=1 \quad \Longrightarrow \quad C=2.
$$

Thus, the decomposition is

$$
\frac{x^2+1}{x^2(x+1)} = -\frac{1}{x}+\frac{1}{x^2}+\frac{2}{x+1}.
$$

**Step 3. Integrate Term by Term**

The integral becomes

$$
\int \left(-\frac{1}{x}+\frac{1}{x^2}+\frac{2}{x+1}\right)dx.
$$

Integrate each term separately:

1. $$\displaystyle \int -\frac{1}{x}\,dx = -\ln|x|.$$

2. $$\displaystyle \int \frac{1}{x^2}\,dx = \int x^{-2}\,dx = -x^{-1} = -\frac{1}{x}.$$

3. $$\displaystyle \int \frac{2}{x+1}\,dx = 2\ln|x+1|.$$

**Step 4. Combine the Results**

Thus, the antiderivative is

$$
-\ln|x| - \frac{1}{x} + 2\ln|x+1| + C,
$$

where $$C$$ is the constant of integration.
The integral of $$ \frac{x^2+1}{x^3+x^2} $$ with respect to $$x$$ is $$ -\ln|x| - \frac{1}{x} + 2\ln|x+1| + C $$, where $$C$$ is the constant of integration.

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