Let I = int tan^{-1} left( frac{2x}{1-x^2} ri | vedclass

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

Question

(B) We know that $	an^{-1} \left( \frac{2x}{1-x^2} ight) = 2 	an^{-1} x$ for $|x| < 1$. Substituting this into the integral,we get $I = \int 2 	a

Vedclass · Accepted Answer

$-\log(1+x^2) + c$

Vedclass · Answer

(B) We know that $	an^{-1} \left( \frac{2x}{1-x^2} ight) = 2 	an^{-1} x$ for $|x| Substituting this into the integral,we get $I = \int 2 	an^{-1} x \, dx$. Using integration by parts,let $u = 	an^{-1} x$ and $dv = 2 \, dx$. Then $du = \frac{1}{1+x^2} \, dx$ and $v = 2x$. Applying the formula $\int u \, dv = uv - \int v \, du$: $I = 2x 	an^{-1} x - \int \frac{2x}{1+x^2} \, dx$. Let $t = 1+x^2$,then $dt = 2x \, dx$. So,$\int \frac{2x}{1+x^2} \, dx = \int \frac{1}{t} \, dt = \log|t| + c = \log(1+x^2) + c$. Thus,$I = 2x 	an^{-1} x - \log(1+x^2) + c$. Rearranging the terms,we get $I - 2x 	an^{-1} x = -\log(1+x^2) + c$.

Let $I = \int \tan^{-1} \left( \frac{2x}{1-x^2} \right) dx$,then $I - 2x \tan^{-1} x = $

Similar Questions

Integrate the function: $x \sin^{-1} x$

$\int x \sin x \sec^3 x \, dx = $

If $I_{m, n} = \int e^{mx} \cdot x^n \, dx$,then $I_{m, n} + \frac{n}{m} I_{m, n-1} =$

$\int (\log x)^2 dx =$

If linear functions $f(x)$ and $g(x)$ satisfy $\int[(3x-1) \cos x + (1-2x) \sin x] dx = f(x) \cos x + g(x) \sin x + C$,then:

Vedclass Test Series

Exam Paper Generator

Online Exam Module