int frac{d x}{(x-2) sqrt{x^2-3 x+5}} = | vedclass

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

Question

(B) माना $I = \int \frac{d x}{(x-2) \sqrt{x^2-3 x+5}}$.$x-2 = \frac{1}{t}$ प्रतिस्थापित करने पर,$dx = -\frac{1}{t^2} dt$ और $x = 2 + \frac{1}{t}$ प्रा

Vedclass · Accepted Answer

$\frac{-1}{\sqrt{3}} \sinh ^{-1}\left[\frac{x+4}{\sqrt{11}(x-2)}\right]+C$

Vedclass · Answer

(B) माना $I = \int \frac{d x}{(x-2) \sqrt{x^2-3 x+5}}$.$x-2 = \frac{1}{t}$ प्रतिस्थापित करने पर,$dx = -\frac{1}{t^2} dt$ और $x = 2 + \frac{1}{t}$ प्राप्त होता है।इन मानों को समाकलन में रखने पर:$I = \int \frac{-1/t^2 dt}{(1/t) \sqrt{(2+1/t)^2 - 3(2+1/t) + 5}}$$I = -\int \frac{dt/t}{\sqrt{4 + 4/t + 1/t^2 - 6 - 3/t + 5}}$$I = -\int \frac{dt}{\sqrt{1/t^2 + 1/t + 3}} = -\int \frac{dt}{\sqrt{(1 + t + 3t^2)/t^2}} = -\int \frac{dt}{\sqrt{3t^2 + t + 1}}$$I = -\frac{1}{\sqrt{3}} \int \frac{dt}{\sqrt{t^2 + t/3 + 1/3}} = -\frac{1}{\sqrt{3}} \int \frac{dt}{\sqrt{(t + 1/6)^2 + 11/36}}$सूत्र $\int \frac{du}{\sqrt{u^2 + a^2}} = \sinh^{-1}(\frac{u}{a}) + C$ का उपयोग करने पर:$I = -\frac{1}{\sqrt{3}} \sinh^{-1}\left( \frac{t + 1/6}{\sqrt{11}/6} \right) + C = -\frac{1}{\sqrt{3}} \sinh^{-1}\left( \frac{6t + 1}{\sqrt{11}} \right) + C$चूंकि $t = \frac{1}{x-2}$,इसलिए:$I = -\frac{1}{\sqrt{3}} \sinh^{-1}\left( \frac{6/(x-2) + 1}{\sqrt{11}} \right) + C = -\frac{1}{\sqrt{3}} \sinh^{-1}\left( \frac{6 + x - 2}{\sqrt{11}(x-2)} \right) + C$$I = -\frac{1}{\sqrt{3}} \sinh^{-1}\left( \frac{x + 4}{\sqrt{11}(x-2)} \right) + C$.

$\int \frac{d x}{(x-2) \sqrt{x^2-3 x+5}} =$

Similar Questions

यदि $\int \frac{5 \tan x}{\tan x-2} dx=ax+b \log |\sin x-2 \cos x|+c$ है,तो $a+b=$

निम्नलिखित समाकल ज्ञात कीजिए: $\int \frac{x+3}{\sqrt{5-4 x-x^{2}}} d x$

$\int \frac{dx}{(1+x) \sqrt{8+7x-x^2}} = $

$\int \sqrt{3-2x-x^2} \, dx = $ . . . . . . $+ C$.

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

Vedclass Test Series

Exam Paper Generator

Online Exam Module