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

If $I_n = \int_{\pi / 2}^{\infty} e^{-x} \cos^n x \, dx$, then $\frac{I_{2018}}{I_{2016}} = $

The value of $I = \int \frac{dx}{x^2(x^4+1)^{3/4}}$ is

If $\int \frac{dx}{\cos^3 x \sqrt{2 \sin 2x}} = (\tan x)^A + C(\tan x)^B + k$ where $k$ is a constant of integration,then $A+B+C$ equals

Integrate the function: $\frac{1}{\sqrt{8+3x-x^{2}}}$

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

Vedclass Test Series

Exam Paper Generator

Online Exam Module