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