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) Let $I = \int \frac{d x}{(x-2) \sqrt{x^2-3 x+5}}$.Substitute $x-2 = \frac{1}{t}$, then $dx = -\frac{1}{t^2} dt$ and $x = 2 + \frac{1}{t}$.Substitu

Vedclass · Accepted Answer

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

Vedclass · Answer

(B) Let $I = \int \frac{d x}{(x-2) \sqrt{x^2-3 x+5}}$.Substitute $x-2 = \frac{1}{t}$, then $dx = -\frac{1}{t^2} dt$ and $x = 2 + \frac{1}{t}$.Substituting these into the integral:$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}}$$I = -\frac{1}{\sqrt{3}} \int \frac{dt}{\sqrt{t^2 + t/3 + 1/3}} = -\frac{1}{\sqrt{3}} \int \frac{dt}{\sqrt{(t + 1/6)^2 + 11/36}}$Using the formula $\int \frac{du}{\sqrt{u^2 + a^2}} = \sinh^{-1}(\frac{u}{a}) + C$:$I = -\frac{1}{\sqrt{3}} \sinh^{-1}\left( \frac{t + 1/6}{\sqrt{11}/6} \right) + C = -\frac{1}{\sqrt{3}} \sinh^{-1}\left( \frac{6t + 1}{\sqrt{11}} \right) + C$Since $t = \frac{1}{x-2}$, we have:$I = -\frac{1}{\sqrt{3}} \sinh^{-1}\left( \frac{6/(x-2) + 1}{\sqrt{11}} \right) + C = -\frac{1}{\sqrt{3}} \sinh^{-1}\left( \frac{6 + x - 2}{\sqrt{11}(x-2)} \right) + C$$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

For $0 < x < 1$,evaluate the integral $\int [\operatorname{Tan}^{-1}(1-x+x^2) + \operatorname{Tan}^{-1}(1-x)] dx$.

If $\int \frac{dx}{x^{2022}(1+x^{2022})^{1/2022}} = \frac{-(1+x^m)^{n/m}}{nx^n} + C$,then $m-n=$

If $\int \frac{dx}{(1+\sqrt{x}) \sqrt{x-x^2}} = \frac{A \sqrt{x}}{\sqrt{1-x}} + \frac{B}{\sqrt{1-x}} + C$, where $C$ is a real constant, then $A+B$ is equal to

For $x>0$, evaluate the integral: $\int \left( \frac{\sqrt{1+x+x^2}}{1+x} + \frac{1}{2 \sqrt{1+x+x^2}} - \frac{1}{(1+x) \sqrt{1+x+x^2}} \right) dx$.

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

Vedclass Test Series

Exam Paper Generator

Online Exam Module