Question

The value of the integral $\int_{0}^{\infty} \frac{\tan^{-1}x}{x^2+1+2x} dx$ is:

• A. $\frac{\pi}{2}$
• B. $\frac{\pi}{4}$
• C. $\frac{\pi}{8}$
• D. $\frac{\pi}{6}$

Answer 1

Answer: (B)

$\frac{\pi}{4}$

Answer 2

The given integral is $I = \int_{0}^{\infty} \frac{\tan^{-1}x}{x^2+1+2x} dx$.

First, simplify the denominator: $x^2+1+2x = (x+1)^2$.
So the integral becomes $I = \int_{0}^{\infty} \frac{\tan^{-1}x}{(x+1)^2} dx$.

We will use integration by parts, $\int u \, dv = uv - \int v \, du$.
Let $u = \tan^{-1}x$ and $dv = \frac{1}{(x+1)^2} dx$.

Then, we find $du$ and $v$:
$du = \frac{1}{1+x^2} dx$
$v = \int \frac{1}{(x+1)^2} dx = -\frac{1}{x+1}$

Substitute these into the integration by parts formula:
$I = \left[ -\frac{\tan^{-1}x}{x+1} \right]_{0}^{\infty} - \int_{0}^{\infty} \left( -\frac{1}{x+1} \right) \left( \frac{1}{1+x^2} \right) dx$

Evaluate the first term:
$\left[ -\frac{\tan^{-1}x}{x+1} \right]_{0}^{\infty} = \lim_{x \to \infty} \left( -\frac{\tan^{-1}x}{x+1} \right) - \left( -\frac{\tan^{-1}0}{0+1} \right)$
As $x \to \infty$, $\tan^{-1}x \to \frac{\pi}{2}$. So, $\lim_{x \to \infty} \left( -\frac{\tan^{-1}x}{x+1} \right) = -\frac{\pi/2}{\infty} = 0$.
Also, $\tan^{-1}0 = 0$. So, $-\frac{\tan^{-1}0}{0+1} = 0$.
Thus, the first term evaluates to $0 - 0 = 0$.

The integral simplifies to:
$I = \int_{0}^{\infty} \frac{1}{(x+1)(x^2+1)} dx$

Now, we use partial fraction decomposition for the integrand:
$\frac{1}{(x+1)(x^2+1)} = \frac{A}{x+1} + \frac{Bx+C}{x^2+1}$
Multiply by $(x+1)(x^2+1)$:
$1 = A(x^2+1) + (Bx+C)(x+1)$

To find $A$, set $x=-1$:
$1 = A((-1)^2+1) + 0 \implies 1 = 2A \implies A = \frac{1}{2}$.

Expand the equation and compare coefficients:
$1 = Ax^2+A + Bx^2+Bx+Cx+C$
$1 = (A+B)x^2 + (B+C)x + (A+C)$

Comparing coefficients of $x^2$: $0 = A+B \implies B = -A = -\frac{1}{2}$.
Comparing constant terms: $1 = A+C \implies C = 1-A = 1-\frac{1}{2} = \frac{1}{2}$.

So, the partial fraction decomposition is:
$\frac{1}{(x+1)(x^2+1)} = \frac{1/2}{x+1} + \frac{-1/2 x + 1/2}{x^2+1} = \frac{1}{2(x+1)} - \frac{x}{2(x^2+1)} + \frac{1}{2(x^2+1)}$

Now, integrate this expression from $0$ to $\infty$:
$I = \int_{0}^{\infty} \left( \frac{1}{2(x+1)} - \frac{x}{2(x^2+1)} + \frac{1}{2(x^2+1)} \right) dx$
$I = \left[ \frac{1}{2} \ln|x+1| - \frac{1}{4} \ln(x^2+1) + \frac{1}{2} \tan^{-1}x \right]_{0}^{\infty}$

Combine the logarithmic terms: $\frac{1}{2} \ln(x+1) - \frac{1}{4} \ln(x^2+1) = \frac{1}{4} [2\ln(x+1) - \ln(x^2+1)] = \frac{1}{4} \ln\left(\frac{(x+1)^2}{x^2+1}\right)$

So, the integral becomes:
$I = \left[ \frac{1}{4} \ln\left(\frac{(x+1)^2}{x^2+1}\right) + \frac{1}{2} \tan^{-1}x \right]_{0}^{\infty}$

Evaluate at the upper limit ($x \to \infty$):
$\lim_{x \to \infty} \left( \frac{1}{4} \ln\left(\frac{(x+1)^2}{x^2+1}\right) + \frac{1}{2} \tan^{-1}x \right)$
For the logarithmic term: $\lim_{x \to \infty} \frac{(x+1)^2}{x^2+1} = \lim_{x \to \infty} \frac{x^2+2x+1}{x^2+1} = \lim_{x \to \infty} \frac{1+2/x+1/x^2}{1+1/x^2} = 1$.
So, $\lim_{x \to \infty} \frac{1}{4} \ln(1) = 0$.
For the $\tan^{-1}x$ term: $\lim_{x \to \infty} \frac{1}{2} \tan^{-1}x = \frac{1}{2} \left( \frac{\pi}{2} \right) = \frac{\pi}{4}$.
Value at upper limit = $0 + \frac{\pi}{4} = \frac{\pi}{4}$.

Evaluate at the lower limit ($x=0$):
$\frac{1}{4} \ln\left(\frac{(0+1)^2}{0^2+1}\right) + \frac{1}{2} \tan^{-1}0$
$= \frac{1}{4} \ln\left(\frac{1}{1}\right) + \frac{1}{2} (0)$
$= \frac{1}{4} \ln(1) + 0 = 0$.

Finally, subtract the lower limit value from the upper limit value:
$I = \frac{\pi}{4} - 0 = \frac{\pi}{4}$.

The final answer is $\frac{\pi}{4}$.