Question Statement

Evaluate the integral:

$\int (1+x)\left(1-x^2\right)dx$

---

Background and Explanation

This problem requires integrating a polynomial function that is given in factored form. The key strategy is to first expand the product into standard polynomial form, then apply the power rule for integration to each term separately.

---

Solution

Let $I=\int (1+x)\left(1-x^2\right)dx$

First, expand the integrand by multiplying $(1+x)$ with $(1-x^2)$:

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

So we can rewrite the integral as:

$I=\int \left(1-x^2+x-x^3\right)dx$

Using the linearity property of integration, we can split this into separate integrals for each term:

$I=\int 1dx-\int x^{2}dx+\int xdx-\int x^{3}dx$

Now apply the power rule $\int x^{n}dx=\frac{x^{n+1}}{n+1}$ to each term:

• $\int 1dx=x$
• $\int x^{2}dx=\frac{x^3}{3}$, so $-\int x^{2}dx=-\frac{x^3}{3}$
• $\int xdx=\frac{x^2}{2}$
• $\int x^{3}dx=\frac{x^4}{4}$, so $-\int x^{3}dx=-\frac{x^4}{4}$

Combining these results and adding the constant of integration $c$:

$I=x-\frac{x^3}{3}+\frac{x^2}{2}-\frac{x^4}{4}+c$

Rearranging the terms in standard polynomial order (ascending or descending powers):

$I=x+\frac{x^2}{2}-\frac{x^3}{3}-\frac{x^4}{4}+c$

---

Key Formulas or Methods Used

• Algebraic expansion: $(a+b)(c+d)=ac+ad+bc+bd$
• Linearity of integration: $\int [f(x)\pm g(x)]dx=\int f(x)dx\pm \int g(x)dx$
• Power rule for integration: $\int x^{n}dx=\frac{x^{n+1}}{n+1}+C$ (for $n\ne -1$)
• Constant of integration: Always add $+c$ for indefinite integrals

---

Summary of Steps

1. Expand the product $(1+x)(1-x^2)$ to get $1-x^2+x-x^3$
2. Split the integral into four separate integrals using linearity
3. Apply the power rule to integrate each term individually:
   • Constant term $\to$ linear term
   • $x$ term $\to$ $\frac{x^2}{2}$
   • $x^2$ term $\to$ $\frac{x^3}{3}$
   • $x^3$ term $\to$ $\frac{x^4}{4}$
4. Combine all terms with appropriate signs and add the constant $c$
5. Rearrange terms in standard order (optional)