Results Summary

If you’re searching for “integration of x 1 x 2,” here are some possible answers:

$$\boxed{\int \frac{x + 1}{x + 2} \, dx = \ln |x+2| – \frac{1}{x+2} + C}$$

$$\boxed{\int \frac{x – 1}{x – 2} \, dx = \ln |x – 2| + \frac{1}{x – 2} + C}$$

$$\boxed{\int \frac{x}{1 + x^2} \, dx = \frac{1}{2} \ln |1 + x^2| + C}$$

$$\boxed{\int x (1 + x)^2 \, dx = \frac{x^2}{2} + \frac{2x^3}{3} + \frac{x^4}{4} + C}$$

Introduction

The “integration of x 1 x 2” could have multiple meanings. Here are four possible options for this expression, with detailed solutions for each.

Option 1: \( \displaystyle \int \frac{x + 1}{x + 2} \, dx \)

Rewrite the integral as:

\[\begin{align*} = \int 1 – \frac{1}{x + 2} \, dx \end{align*}\]

Now integrate each term:

\[\begin{align*} = x – \ln |x + 2| + C \end{align*}\]

Thus, the result is:

\[\begin{align*} \int \frac{x + 1}{x + 2} \, dx = x – \ln |x + 2| + C \end{align*}\]

Option 2: \( \displaystyle \int \frac{x – 1}{x – 2} \, dx \)

Rewrite the integral as:

\[\begin{align*} = \int 1 + \frac{1}{x – 2} \, dx \end{align*}\]

Now integrate each term:

\[\begin{align*} = x + \ln |x – 2| + C \end{align*}\]

Thus, the result is:

\[\begin{align*} \int \frac{x – 1}{x – 2} \, dx = x + \ln |x – 2| + C \end{align*}\]

Option 3: \( \displaystyle \int \frac{x}{1 + x^2} \, dx \)

Using the $u$-substitution \( u = 1 + x^2 \), so \( du = 2x \, dx \) or \( \frac{du}{2} = x \, dx \):

\[\begin{align*} = \int \frac{1}{2} \cdot \frac{1}{u} \, du = \frac{1}{2} \ln |u| + C \end{align*}\]

Substitute back \( u = 1 + x^2 \):

\[\begin{align*} = \frac{1}{2} \ln |1 + x^2| + C \end{align*}\]

Thus, the result is:

\[\begin{align*} \int \frac{x}{1 + x^2} \, dx = \frac{1}{2} \ln |1 + x^2| + C \end{align*}\]

Option 4: \( \displaystyle\int x (1 + x)^2 \, dx \)

Expanding \( (1 + x)^2 = 1 + 2x + x^2 \):

\[\begin{align*} = \int x (1 + 2x + x^2) \, dx = \int (x + 2x^2 + x^3) \, dx \end{align*}\]

Now integrate each term:

\[\begin{align*} = \frac{x^2}{2} + \frac{2x^3}{3} + \frac{x^4}{4} + C \end{align*}\]

Thus, the result is:

\[\begin{align*} \int x (1 + x)^2 \, dx = \frac{x^2}{2} + \frac{2x^3}{3} + \frac{x^4}{4} + C \end{align*}\]