Results Summary

If you’re searching for “integral of 1 1 x 2,” here are some possible interpretations and their solutions:

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

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

Let’s go through each interpretation and solve it step-by-step.

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

This is a standard integral that results in the arctangent function:

\[\begin{align*} = \arctan(x) + C \end{align*}\]

Thus, the result is:

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

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

This integral can be evaluated by recognizing it as a partial fraction decomposition. Rewrite \( \dfrac{1}{1 – x^2} \) as:

\[\begin{align*} \frac{1}{1 – x^2} = \frac{1}{2} \left( \frac{1}{1 + x} + \frac{1}{1 – x} \right) \end{align*}\]

Then integrate each term:

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

Combining the logarithmic terms, we get:

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

Thus, the result is:

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