Question

The integral $\int x \,cos^{-1}\left(\frac{1-x^{2}}{1+x^{2}}\right) \,dx\, \left(x > 0\right)$ is equal to :

• A. $- x + (1 + x^2) tan^{-1} x + c$
• B. $x- (1 + x^2) cot^{-1} x + c$
• C. $- x + (1 + x^2) cot^{-1} x + c$
• D. $x- (1 + x^2) tan^{-1} x + c$

Answer 1

The Correct Option is A

To solve the integral \(\int x \, \cos^{-1}\left(\frac{1-x^{2}}{1+x^{2}}\right) \, dx\, (x > 0)\), we start by exploring the expression for the inverse cosine function. 

First, let's analyze the argument of \(\cos^{-1}\), which is \(\frac{1-x^{2}}{1+x^{2}}\). Notice that this is a well-known identity related to the tangent inverse function:

Recall that:

• \(\tan \theta = \frac{2x}{1-x^2}\) for some angle \(\theta\) such that \(\cos \theta = \frac{1-x^{2}}{1+x^{2}}\).
• This means \(\cos^{-1}\left(\frac{1-x^{2}}{1+x^{2}}\right) = 2 \tan^{-1} x\).

Hence, transforming the original integral becomes essential:

The integral transforms to:

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

Now we proceed to solve the integral \(2 \int x \, \tan^{-1} x \, dx\) using integration by parts, where:

• Let \(u = \tan^{-1} x \implies du = \frac{1}{1+x^2} \, dx\)
• Let \(dv = x \, dx \implies v = \frac{x^2}{2}\)

Now apply the integration by parts formula:

\[\int u \, dv = uv - \int v \, du\]

Substitute back into the formula:

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

Simplify the integral:

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

Therefore, integrating simplifies to:

\[= (1 + x^2) \tan^{-1} x - x\]

Thus, including the integration constant, we get:

The solution to the integral is:

\[- x + (1 + x^2) \tan^{-1} x + c\]

This matches the given correct option, confirming the answer. The correct option is:

\[- x + (1 + x^2) \tan^{-1} x + c\]