Evaluate :
`int tan^(-1)((x-5)/(1+5x))dx`

Answer

To evaluate the integral \( I = \int \tan^{-1}\left(\frac{x-5}{1+5x}\right)dx \), we can use a trigonometric identity for the inverse tangent function.

### Step 1: Simplify the integrand
Using the identity for the tangent of the difference of two angles, we have:
\[ \tan^{-1}(a) - \tan^{-1}(b) = \tan^{-1\left(\frac{a-b}{1+ab}\right)} \]
In our case, let \( a = x \) and \( b = 5 \). Then:
\[ \tan^{-1}\left(\frac{x-5}{1+5x}\right) = \tan^{-1}(x) - \tan^{-1}(5) \]

### Step 2: Rewrite the integral
Now we can rewrite the integral as:
\[ I = \int \left(\tan^{-1}(x) - \tan^{-1}(5)\right)dx \]
This can be split into two separate integrals:
\[ I = \int \tan^{-1}(x)dx - \int \tan^{-1}(5)dx \]

### Step 3: Evaluate the second integral
The second integral is straightforward since \(\tan^{-1}(5)\) is a constant:
\[ \int \tan^{-1}(5)dx = x \tan^{-1}(5) \]

### Step 4: Evaluate the first integral
Now we need to evaluate \( \int \tan^{-1}(x)dx \). We will use integration by parts. Let:
- \( u = \tan^{-1}(x) \) so that \( du = \frac{1}{1+x^2}dx \)
- \( dv = dx \) so that \( v = x \)

Now applying integration by parts:
\[ \int u \, dv = uv - \int v \, du \]
This gives us:
\[ \int \tan^{-1}(x)dx = x \tan^{-1}(x) - \int \frac{x}{1+x^2}dx \]

### Step 5: Evaluate the remaining integral
The remaining integral can be evaluated as follows:
\[ \int \frac{x}{1+x^2}dx \]
Using the substitution \( w = 1+x^2 \), \( dw = 2x \, dx \) or \( dx = \frac{dw}{2x} \):
\[ \int \frac{x}{1+x^2}dx = \frac{1}{2} \int \frac{1}{w}dw = \frac{1}{2} \ln|w| + C = \frac{1}{2} \ln(1+x^2) + C \]

### Step 6: Combine results
Putting everything together, we have:
\[ \int \tan^{-1}(x)dx = x \tan^{-1}(x) - \frac{1}{2} \ln(1+x^2) + C \]
Now substituting back into our expression for \( I \):
\[ I = \left(x \tan^{-1}(x) - \frac{1}{2} \ln(1+x^2)\right) - \left(x \tan^{-1}(5)\right) + C \]
Thus, we can simplify this to:
\[ I = x \tan^{-1}(x) - x \tan^{-1}(5) - \frac{1}{2} \ln(1+x^2) + C \]

### Final Answer
\[ I = x \tan^{-1}(x) - x \tan^{-1}(5) - \frac{1}{2} \ln(1+x^2) + C \]