Question

Evaluate the integral: ∫ 11 tan⁻¹(x)/(x)² dx

Answer 1

The evaluation of the integral
`\(\int \frac{{\tan^(-1)(x)}}{{x^2}} \,dx\)` results in
`\(\frac{{\tan^(-2)(x)}}{2} + C\), where \(C\)` is the constant of integration.

Step-by-step explanation:

To evaluate the given integral, we can use integration by parts. Let
`\(u = \tan^(-1)(x)\) and \(dv = \frac{1}{{x^2}} \,dx\).` Then, we differentiate
`\(u\) to get \(du = \frac{1}{{1 + x^2}} \,dx\)` and integrate
`\(dv\) to get \(v = -(1)/(x)\).` Applying the integration by parts formula:

`\[ \int \frac{{\tan^(-1)(x)}}{{x^2}} \,dx = uv - \int v \,du \]`

`\[ = \tan^(-1)(x) \left(-(1)/(x)\right) - \int \left(-(1)/(x)\right) \left(\frac{1}{{1 + x^2}} \,dx\right) \]`

Simplifying further:

`\[ = -(\tan^(-1)(x))/(x) + \int \frac{1}{{x(1 + x^2)}} \,dx \]`

Now, to evaluate the remaining integral, we can use partial fraction decomposition or another appropriate method. Once integrated, we arrive at the final result:

`\[ = -(\tan^(-1)(x))/(x) + (\tan^(-1)(x))/(2) + C \]`

where
`\(C\)` is the constant of integration. Thus, the integral
`\(\int \frac{{\tan^(-1)(x)}}{{x^2}} \,dx\) is evaluated as \(\frac{{\tan^(-2)(x)}}{2} + C\).`