int_0^1 x tan^{-1} x , dx = | vedclass

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

Question

(B) Using integration by parts: $\int u \, dv = uv - \int v \, du$. Let $u = 	an^{-1} x$ and $dv = x \, dx$. Then $du = \frac{1}{1+x^2} \, dx$ and $v

Vedclass · Accepted Answer

$\frac{\pi}{4} - \frac{1}{2}$

Vedclass · Answer

(B) Using integration by parts: $\int u \, dv = uv - \int v \, du$. Let $u = 	an^{-1} x$ and $dv = x \, dx$. Then $du = \frac{1}{1+x^2} \, dx$ and $v = \frac{x^2}{2}$.$\int_0^1 x 	an^{-1} x \, dx = \left[ \frac{x^2}{2} 	an^{-1} x ight]_0^1 - \int_0^1 \frac{x^2}{2(1+x^2)} \, dx$$= \left( \frac{1}{2} \cdot \frac{\pi}{4} - 0 ight) - \frac{1}{2} \int_0^1 \frac{x^2+1-1}{1+x^2} \, dx$$= \frac{\pi}{8} - \frac{1}{2} \int_0^1 \left( 1 - \frac{1}{1+x^2} ight) \, dx$$= \frac{\pi}{8} - \frac{1}{2} \left[ x - 	an^{-1} x ight]_0^1$$= \frac{\pi}{8} - \frac{1}{2} \left( (1 - \frac{\pi}{4}) - (0 - 0) ight)$$= \frac{\pi}{8} - \frac{1}{2} + \frac{\pi}{8} = \frac{\pi}{4} - \frac{1}{2}$.

$\int_0^1 x \tan^{-1} x \, dx = $

Similar Questions

The value of the definite integral,$\int\limits_0^{\frac{\pi }{2}} {\frac{{\sin 5x}}{{\sin x}}\,dx} $ is

$\int_0^{\pi /2} \frac{x + \sin x}{1 + \cos x} \, dx = $

$\int_{e^{-1}}^{e^2} \left| \frac{\log x}{x} \right| dx =$

Let $[x]$ and $\{x\}$ be the integer part and fractional part of a real number $x$ respectively. The value of the integral $\int_0^5 [x]\{x\} dx$ is

$\mathop {Lim}\limits_{n \to \infty } \int_0^2 {\left( {1 + \frac{t}{{n + 1}}} \right)^n} dt$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module