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

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

Question

(B) We need to evaluate the integral $I = \int \frac{dx}{\sqrt{(x-1)(x-2)}}$.First,expand the denominator: $(x-1)(x-2) = x^2 - 3x + 2$.Now,complete th

Vedclass · Accepted Answer

$\log \left|\left(x-\frac{3}{2}\right)+\sqrt{x^{2}-3 x+2}\right|+c$

Vedclass · Answer

(B) We need to evaluate the integral $I = \int \frac{dx}{\sqrt{(x-1)(x-2)}}$.First,expand the denominator: $(x-1)(x-2) = x^2 - 3x + 2$.Now,complete the square for the quadratic expression: $x^2 - 3x + 2 = (x^2 - 3x + \frac{9}{4}) - \frac{9}{4} + 2 = (x - \frac{3}{2})^2 - \frac{1}{4} = (x - \frac{3}{2})^2 - (\frac{1}{2})^2$.The integral becomes $I = \int \frac{dx}{\sqrt{(x - \frac{3}{2})^2 - (\frac{1}{2})^2}}$.Using the standard integral formula $\int \frac{dx}{\sqrt{x^2 - a^2}} = \log |x + \sqrt{x^2 - a^2}| + c$,we get:$I = \log |(x - \frac{3}{2}) + \sqrt{(x - \frac{3}{2})^2 - (\frac{1}{2})^2}| + c$.Simplifying the expression inside the square root back to the original form,we get:$I = \log |(x - \frac{3}{2}) + \sqrt{x^2 - 3x + 2}| + c$.

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

Similar Questions

Integrate the function $\frac{1}{1+\cot x}$.

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

If $\int \frac{1-(\cot x)^{2019}}{\tan x+(\cot x)^{2020}} dx = \frac{1}{n} \ln |(f(x))^n + (g(x))^n| + c$,then the value of $n[(f(x))^4 + (g(x))^4]_{x=\frac{\pi}{3}}$ is:

$\int \frac{13 \cos 2 x-9 \sin 2 x}{3 \cos 2 x-4 \sin 2 x} d x=$

If $\int \sin (101 x)(\sin x)^{99} d x=\frac{\sin (100 x)(\sin x)^\lambda}{\mu}+c$ then $\frac{\lambda}{\mu}=$

Vedclass Test Series

Exam Paper Generator

Online Exam Module