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