The integral int_0^1 cot^{-1}(1 + x + x^2) dx | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) We know that $\cot^{-1} x = 	an^{-1} \left(\frac{1}{x}ight)$ for $x > 0$.
Thus, $\cot^{-1}(1 + x + x^2) = 	an^{-1}\left(\frac{1}{1+x(x+1)}

Vedclass · Accepted Answer

$2 	an^{-1} 2 - \frac{1}{2} \log_e \left(\frac{5}{4}ight) - \frac{\pi}{2}$

Vedclass · Answer

(D) We know that $\cot^{-1} x = 	an^{-1} \left(\frac{1}{x}ight)$ for $x > 0$.
Thus, $\cot^{-1}(1 + x + x^2) = 	an^{-1}\left(\frac{1}{1+x(x+1)}ight) = 	an^{-1}(x+1) - 	an^{-1}x$.
Integrating by parts, $\int 	an^{-1} u \, du = u 	an^{-1} u - \frac{1}{2} \log_e(1+u^2) + C$.
Applying this to the integral $I = \int_0^1 	an^{-1}(x+1) dx - \int_0^1 	an^{-1} x dx$:
$I = \left[ (x+1) 	an^{-1}(x+1) - \frac{1}{2} \log_e(1+(x+1)^2) ight]_0^1 - \left[ x 	an^{-1} x - \frac{1}{2} \log_e(1+x^2) ight]_0^1$.
Evaluating at the limits:
$I = \left( 2 	an^{-1} 2 - \frac{1}{2} \log_e 5 - (1 	an^{-1} 1 - \frac{1}{2} \log_e 2) ight) - \left( 1 	an^{-1} 1 - \frac{1}{2} \log_e 2 - 0 ight)$.
$I = 2 	an^{-1} 2 - \frac{1}{2} \log_e 5 - 2 	an^{-1} 1 + \log_e 2$.
Since $	an^{-1} 1 = \frac{\pi}{4}$, we have $I = 2 	an^{-1} 2 - \frac{\pi}{2} + \frac{1}{2} \log_e \left(\frac{4}{5}ight) = 2 	an^{-1} 2 - \frac{\pi}{2} - \frac{1}{2} \log_e \left(\frac{5}{4}ight)$.

The integral $\int_0^1 \cot^{-1}(1 + x + x^2) dx$ is equal to:

Similar Questions

If $f(t) = \int_{-t}^t \frac{e^{-|x|}}{2} dx$, then $\lim_{t \rightarrow \infty} f(t)$ is equal to:

If $[x]$ is the greatest integer function not greater than $x$,then $\int_0^8 [x] dx$ is equal to

If $g(1) = g(2)$,then $\int_1^2 {{{\left[ {f(g(x))} \right]}^{ - 1}}} f'\{ g(x)\} \;g'(x)\;dx$ is equal to

The value of $\int_{0}^{1} \frac{x^{4}(1-x)^{4}}{1+x^{2}} dx$ is equal to

The value of the definite integral $\int_{1}^{\infty} (e^{x+1} + e^{3-x})^{-1} \, dx$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module