int tan ^{-1}left(1-x+x^2right) d x+int tan ^ | vedclass

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

Question

(A) Let $I = \int 	an ^{-1}(1-x+x^2) dx + \int 	an ^{-1}(x) dx + \int 	an ^{-1}(1-x) dx$. Using the property $	an^{-1}(A) + 	an^{-1}(B) = 	an^{-

Vedclass · Accepted Answer

$\frac{\pi}{2} x+C$

Vedclass · Answer

(A) Let $I = \int 	an ^{-1}(1-x+x^2) dx + \int 	an ^{-1}(x) dx + \int 	an ^{-1}(1-x) dx$. Using the property $	an^{-1}(A) + 	an^{-1}(B) = 	an^{-1}\left(\frac{A+B}{1-AB}ight)$,we combine the last two terms: $	an^{-1}(x) + 	an^{-1}(1-x) = 	an^{-1}\left(\frac{x + 1 - x}{1 - x(1-x)}ight) = 	an^{-1}\left(\frac{1}{1-x+x^2}ight)$. Since $	an^{-1}\left(\frac{1}{u}ight) = \cot^{-1}(u)$ for $u > 0$,we have: $I = \int \left[ 	an^{-1}(1-x+x^2) + \cot^{-1}(1-x+x^2) ight] dx$. Using the identity $	an^{-1}(u) + \cot^{-1}(u) = \frac{\pi}{2}$: $I = \int \frac{\pi}{2} dx = \frac{\pi}{2} x + C$.

$\int \tan ^{-1}\left(1-x+x^2\right) d x+\int \tan ^{-1}(x) d x+\int \tan ^{-1}(1-x) d x=$

Similar Questions

Let $\alpha \in (0, \frac{\pi}{2})$ be fixed. If the integral $\int \frac{\tan x+\tan \alpha}{\tan x-\tan \alpha} dx = A(x) \cos 2\alpha + B(x) \sin 2\alpha + c$ (where $c$ is a constant of integration),then functions $A(x)$ and $B(x)$ are respectively

Evaluate the integral $\int \frac{1}{\sqrt{(x-1)(x-2)}} dx$.

$\int \frac{dx}{(2\sin x + \cos x)^2} = $

If $\int \frac{\sin ^8 x-\cos ^8 x}{1-2 \sin ^2 x \cos ^2 x} d x=A \sin 2 x+B$,then $A$ is equal to

If $\int \frac{e^{\frac{x}{2}}}{\sqrt{e^{-x}-e^x}} \, dx = \sin^{-1}(f(x)) + C$,(where $C$ is the constant of integration),then $f(2)$ has the value:

Vedclass Test Series

Exam Paper Generator

Online Exam Module