int frac{dx}{(x+1) sqrt{x^2+1}} = | vedclass

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

Question

(C) ધારો કે $I = \int \frac{dx}{(x+1) \sqrt{x^2+1}}$.$x+1 = \frac{1}{t}$ આદેશ લેતા,$dx = -\frac{1}{t^2} dt$ મળે.તેથી $x = \frac{1}{t} - 1 = \frac{1-t}

Vedclass · Accepted Answer

$-\frac{1}{\sqrt{2}} \operatorname{Sinh}^{-1}\left(\frac{1+x}{\sqrt{2}(1-x)}\right) + c$

Vedclass · Answer

(C) ધારો કે $I = \int \frac{dx}{(x+1) \sqrt{x^2+1}}$.$x+1 = \frac{1}{t}$ આદેશ લેતા,$dx = -\frac{1}{t^2} dt$ મળે.તેથી $x = \frac{1}{t} - 1 = \frac{1-t}{t}$.હવે $x^2+1 = \left(\frac{1-t}{t}\right)^2 + 1 = \frac{2t^2-2t+1}{t^2}$.આ કિંમતો સંકલનમાં મૂકતા:$I = \int \frac{-dt/t^2}{(1/t) \sqrt{(2t^2-2t+1)/t^2}} = -\int \frac{dt}{\sqrt{2t^2-2t+1}} = -\frac{1}{\sqrt{2}} \int \frac{dt}{\sqrt{(t-1/2)^2 + 1/4}}$.સૂત્ર $\int \frac{du}{\sqrt{u^2+a^2}} = \operatorname{Sinh}^{-1}(\frac{u}{a})$ નો ઉપયોગ કરતા:$I = -\frac{1}{\sqrt{2}} \operatorname{Sinh}^{-1}\left(\frac{t-1/2}{1/2}\right) + c = -\frac{1}{\sqrt{2}} \operatorname{Sinh}^{-1}(2t-1) + c$.$t = \frac{1}{x+1}$ હોવાથી,$2t-1 = \frac{1-x}{x+1}$.તેથી,$I = -\frac{1}{\sqrt{2}} \operatorname{Sinh}^{-1}\left(\frac{1-x}{1+x}\right) + c$.

$\int \frac{dx}{(x+1) \sqrt{x^2+1}} = $

Similar Questions

$\int \left[ \log(\log x) + \frac{1}{(\log x)^2} \right] dx$ શોધો.

$\int \frac{\cos^3 x + \cos^5 x}{\sin^2 x + \sin^4 x} dx =$

જો $\int \frac{7 x^8+8 x^7}{\left(1+x+x^8\right)^2} d x=f(x)+c$ હોય,તો $f(x)$ ની કિંમત શોધો.

જો $\int \sec ^2 x \operatorname{cosec}^4 x \, dx = -\frac{1}{3} \cot ^3 x + k \tan x - 2 \cot x + C$ હોય,તો $k$ ની કિંમત શોધો.

$\int \frac{x - 1}{(x + 1) \sqrt{x(x^2 + x + 1)}} dx =$

Vedclass Test Series

Exam Paper Generator

Online Exam Module