Answer :
Sure, let's solve the indefinite integral \(\int \frac{a}{n^2 + x^2} \, dx\).
1. Identify the integrand:
The integrand is \(\frac{a}{n^2 + x^2}\).
2. Recall the standard integral:
The standard integral for \(\int \frac{1}{x^2 + a^2} \, dx\) is \(\frac{1}{a} \arctan \left( \frac{x}{a} \right) + C\).
3. Apply the standard result:
Here, our integrand is of a similar form, but with an extra constant \(a\) multiplied in the numerator.
[tex]\[ \int \frac{a}{n^2 + x^2} \, dx = a \int \frac{1}{n^2 + x^2} \, dx. \][/tex]
4. Adjust the integral accordingly:
We need to match this to the standard result. Note that in our case, \(a^2\) in the standard form corresponds to \(n^2\). Thus, \(a = n\).
This leads to:
[tex]\[ a \int \frac{1}{n^2 + x^2} \, dx = a \cdot \frac{1}{n} \arctan\left(\frac{x}{n}\right) + C. \][/tex]
5. Simplify the expression:
[tex]\[ a \cdot \frac{1}{n} \arctan \left( \frac{x}{n} \right) + C = \frac{a}{n} \arctan \left( \frac{x}{n} \right) + C. \][/tex]
However, upon comparing this step-wise derived solution with the given definite result:
[tex]\[ a \left( - \frac{i}{2} \log ( -i n + x) + \frac{i}{2} \log (i n + x) \right) / n \][/tex]
We can understand that our integrand can also be expressed in terms of complex logarithms due to the presence of imaginary \(i\) in the result.
To summarize,
[tex]\[ \int \frac{a}{n^2 + x^2} \, dx = a \left( - \frac{i}{2} \log ( -i n + x) + \frac{i}{2} \log (i n + x) \right) / n + C \][/tex]
Thus, the more detailed, oriented solution confirms we match the given derived result.
1. Identify the integrand:
The integrand is \(\frac{a}{n^2 + x^2}\).
2. Recall the standard integral:
The standard integral for \(\int \frac{1}{x^2 + a^2} \, dx\) is \(\frac{1}{a} \arctan \left( \frac{x}{a} \right) + C\).
3. Apply the standard result:
Here, our integrand is of a similar form, but with an extra constant \(a\) multiplied in the numerator.
[tex]\[ \int \frac{a}{n^2 + x^2} \, dx = a \int \frac{1}{n^2 + x^2} \, dx. \][/tex]
4. Adjust the integral accordingly:
We need to match this to the standard result. Note that in our case, \(a^2\) in the standard form corresponds to \(n^2\). Thus, \(a = n\).
This leads to:
[tex]\[ a \int \frac{1}{n^2 + x^2} \, dx = a \cdot \frac{1}{n} \arctan\left(\frac{x}{n}\right) + C. \][/tex]
5. Simplify the expression:
[tex]\[ a \cdot \frac{1}{n} \arctan \left( \frac{x}{n} \right) + C = \frac{a}{n} \arctan \left( \frac{x}{n} \right) + C. \][/tex]
However, upon comparing this step-wise derived solution with the given definite result:
[tex]\[ a \left( - \frac{i}{2} \log ( -i n + x) + \frac{i}{2} \log (i n + x) \right) / n \][/tex]
We can understand that our integrand can also be expressed in terms of complex logarithms due to the presence of imaginary \(i\) in the result.
To summarize,
[tex]\[ \int \frac{a}{n^2 + x^2} \, dx = a \left( - \frac{i}{2} \log ( -i n + x) + \frac{i}{2} \log (i n + x) \right) / n + C \][/tex]
Thus, the more detailed, oriented solution confirms we match the given derived result.
