Li Coleman
05/22/2023 · Senior High School
\( \int - \frac { 1 } { \sqrt { x ^ { 2 } + a ^ { 2 } } } d x \)
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Tutor-Verified Answer
Step-by-step Solution
To solve the integral
\[
\int -\frac{1}{\sqrt{x^2 + a^2}} \, dx,
\]
we can first rewrite it as
\[
-\int \frac{1}{\sqrt{x^2 + a^2}} \, dx.
\]
The integral
\[
\int \frac{1}{\sqrt{x^2 + a^2}} \, dx
\]
is a standard integral. The result of this integral is
\[
\ln |x + \sqrt{x^2 + a^2}| + C,
\]
where \( C \) is the constant of integration.
Thus, we have:
\[
-\int \frac{1}{\sqrt{x^2 + a^2}} \, dx = -\left( \ln |x + \sqrt{x^2 + a^2}| + C \right).
\]
This simplifies to:
\[
-\ln |x + \sqrt{x^2 + a^2}| - C.
\]
We can combine the constant terms into a single constant, so we can write the final result as:
\[
-\ln |x + \sqrt{x^2 + a^2}| + C,
\]
where \( C \) is a new constant of integration.
Therefore, the final answer is:
\[
\int -\frac{1}{\sqrt{x^2 + a^2}} \, dx = -\ln |x + \sqrt{x^2 + a^2}| + C.
\]
Quick Answer
\[
\int -\frac{1}{\sqrt{x^2 + a^2}} \, dx = -\ln |x + \sqrt{x^2 + a^2}| + C.
\]
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