Question

$\int \frac{x\text{Tan}^{-1}x}{(1+x^2)^2}dx =$

• A. $\frac{(\text{Tan}^{-1}x)^2}{4} + \frac{x\text{Tan}^{-1}x}{2(1+x^2)} - \frac{1-x^2}{4(1+x^2)} + c$
• B. $\frac{(\text{Tan}^{-1}x)^2}{4} + \frac{4x\text{Tan}^{-1}x+1-x^2}{8(1+x^2)} + c$
• C. $\frac{(\text{Tan}^{-1}x)^2}{4} - \frac{x\text{Tan}^{-1}x}{1+x^2} + \frac{1-x^2}{4(1+x^2)} + c$
• D. $\frac{(\tan x)^2}{4} + \frac{4x\text{Tan}^{-1}x-1+x^2}{4(1+x^2)} + c$
• E. None of these

Hint:

For integrals involving $\tan^{-1}x$ and $(1+x^2)$, the substitution $x=\tan\theta$ is extremely powerful. It simplifies the algebraic structure into a trigonometric integral, which can often be solved using standard techniques like integration by parts or trigonometric identities.

Answer 1

Answer: (E)

Step 1: Understanding the Concept
This integral requires a strategic choice of method, such as integration by parts or trigonometric substitution. Given the structure of the integrand, trigonometric substitution seems promising.
Step 2: Key Formula or Approach
1. Use the substitution $x = \tan\theta$. This implies $dx = \sec^2\theta d\theta$ and $\tan^{-1}x = \theta$. 2. The term $1+x^2$ becomes $1+\tan^2\theta = \sec^2\theta$. 3. Substitute these into the integral to get an integral in terms of $\theta$. 4. Solve the new integral, likely using integration by parts. 5. Substitute back to express the result in terms of $x$.
Step 3: Detailed Explanation
Let $I = \int \frac{x\tan^{-1}x}{(1+x^2)^2}dx$. 1. Substitution: Let $x=\tan\theta$, so $dx = \sec^2\theta d\theta$. \[ I = \int \frac{\tan\theta \cdot \theta}{(1+\tan^2\theta)^2} \sec^2\theta d\theta \] \[ I = \int \frac{\theta\tan\theta}{(\sec^2\theta)^2} \sec^2\theta d\theta = \int \frac{\theta\tan\theta}{\sec^2\theta} d\theta \] \[ I = \int \theta \frac{\sin\theta}{\cos\theta} \cos^2\theta d\theta = \int \theta \sin\theta \cos\theta d\theta \] Using the identity $\sin(2\theta)=2\sin\theta\cos\theta$: \[ I = \frac{1}{2} \int \theta \sin(2\theta) d\theta \] 2. Integration by Parts: Let $u = \theta$ and $dv = \sin(2\theta)d\theta$. Then $du=d\theta$ and $v = \int \sin(2\theta)d\theta = -\frac{1}{2}\cos(2\theta)$. Using the formula $\int u dv = uv - \int v du$: \[ I = \frac{1}{2} \left[ \theta \left(-\frac{1}{2}\cos(2\theta)\right) - \int \left(-\frac{1}{2}\cos(2\theta)\right) d\theta \right] \] \[ I = \frac{1}{2} \left[ -\frac{\theta}{2}\cos(2\theta) + \frac{1}{2} \int \cos(2\theta) d\theta \right] \] \[ I = \frac{1}{2} \left[ -\frac{\theta}{2}\cos(2\theta) + \frac{1}{2} \left(\frac{1}{2}\sin(2\theta)\right) \right] + C \] \[ I = -\frac{\theta}{4}\cos(2\theta) + \frac{1}{8}\sin(2\theta) + C \] 3. Substitute back to x: We have $\theta = \tan^{-1}x$. We also need $\cos(2\theta)$ and $\sin(2\theta)$ in terms of $x$. \[ \cos(2\theta) = \frac{1-\tan^2\theta}{1+\tan^2\theta} = \frac{1-x^2}{1+x^2} \] \[ \sin(2\theta) = \frac{2\tan\theta}{1+\tan^2\theta} = \frac{2x}{1+x^2} \] Substituting these back into the expression for $I$: \[ I = -\frac{\tan^{-1}x}{4} \left(\frac{1-x^2}{1+x^2}\right) + \frac{1}{8} \left(\frac{2x}{1+x^2}\right) + C \] \[ I = \frac{x}{4(1+x^2)} - \frac{(1-x^2)\tan^{-1}x}{4(1+x^2)} + C = \frac{x - (1-x^2)\tan^{-1}x}{4(1+x^2)} + C \] This result is correct, but it does not match any of the complex forms given in the options. The provided options likely correspond to a different, but similar-looking, integral. Step 4: Final Answer
The calculated result for the integral is $\frac{x - (1-x^2)\tan^{-1}x}{4(1+x^2)} + C$. As this does not match any of the provided choices, the question or options are likely flawed.