Results Summary

If you’re searching for “integrate 2 1 x,” here are some possible solutions:

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

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

$$\boxed{\int \left( 2 + \frac{1}{x} \right) \, dx = 2x + \ln |x| + C}$$

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

Introduction

Let’s solve each option step-by-step to make sure we cover all possibilities.

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

Using the rule \( \int \frac{1}{u} \, du = \ln |u| \), where \( u = 1 + x \) and \( du = dx \), we get:

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

Thus, the result is:

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

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

We are going to use $u$-substitution. Let \( u = 1 – x \), so \( du = -dx \). Substituting and integrating:

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

Thus, the result is:

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

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

This integral separates into two integrals:

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

Solving each term:

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

Thus, the result is:

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

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

Let’s split the integral into two simpler integrals:

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

Solving each term:

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

Thus, the result is:

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