Question

The value of $\int x \tan^{-1}x \, dx$ is given by}

• A. $\frac{x^2}{2}\cot^{-1}x - \frac{1}{2}x - \frac{1}{2}\cot^{-1}x + C$
• B. $\frac{x^2}{2}\sec^{-1}x - \frac{1}{2}x - \frac{1}{2}\sec^{-1}x + C$
• C. $\frac{x^2}{2}\tan^{-1}x - \frac{1}{2}x + \frac{1}{2}\tan^{-1}x + C$
• D. $\frac{x^2}{2}\tan^{-1}x - \frac{1}{2}x - \frac{1}{2}\tan^{-1}x + C$

Hint:

Adding and subtracting $1$ in the numerator of $\frac{x^2}{1+x^2}$ is a standard algebraic trick that avoids polynomial long division and simplifies the integrand instantly.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
To integrate a product of an algebraic function and an inverse trigonometric function, we apply the integration by parts method.
Key Formula or Approach:
The integration by parts formula is: \[ \int u \, dv = uv - \int v \, du \] According to the ILATE rule, we choose the inverse trigonometric function as $u$ and the algebraic function as $dv$.

Step 2: Detailed Explanation:
Let: \[ u = \tan^{-1}x \implies du = \frac{1}{1+x^2} \, dx \] \[ dv = x \, dx \implies v = \frac{x^2}{2} \]
Applying the formula: \[ \int x \tan^{-1}x \, dx = \frac{x^2}{2}\tan^{-1}x - \int \frac{x^2}{2(1+x^2)} \, dx \] \[ = \frac{x^2}{2}\tan^{-1}x - \frac{1}{2} \int \frac{x^2}{1+x^2} \, dx \]
To integrate $\frac{x^2}{1+x^2}$, we add and subtract 1 in the numerator: \[ \int \frac{x^2}{1+x^2} \, dx = \int \frac{x^2 + 1 - 1}{1+x^2} \, dx = \int \left( 1 - \frac{1}{1+x^2} \right) \, dx \] \[ = x - \tan^{-1}x \]
Now, substitute this result back into our primary equation: \[ \int x \tan^{-1}x \, dx = \frac{x^2}{2}\tan^{-1}x - \frac{1}{2} \left( x - \tan^{-1}x \right) + C \] \[ = \frac{x^2}{2}\tan^{-1}x - \frac{1}{2}x + \frac{1}{2}\tan^{-1}x + C \]

Step 3: Final Answer:
The value of the integral is $\frac{x^2}{2}\tan^{-1}x - \frac{1}{2}x + \frac{1}{2}\tan^{-1}x + C$.