Evaluate the integral int frac{1}{sqrt{(x-1)( | vedclass

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

Question

(C) We need to evaluate $I = \int \frac{1}{\sqrt{(x-1)(x-2)}} dx$. First,expand the expression inside the square root: $(x-1)(x-2) = x^2 - 3x + 2$. Co

Vedclass · Accepted Answer

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

Vedclass · Answer

(C) We need to evaluate $I = \int \frac{1}{\sqrt{(x-1)(x-2)}} dx$. First,expand the expression inside the square root: $(x-1)(x-2) = x^2 - 3x + 2$. Complete the square for the quadratic expression: $x^2 - 3x + 2 = (x^2 - 3x + \frac{9}{4}) - \frac{9}{4} + 2 = (x - \frac{3}{2})^2 - \frac{1}{4}$. Now the integral becomes $I = \int \frac{1}{\sqrt{(x - \frac{3}{2})^2 - (\frac{1}{2})^2}} dx$. Using the standard formula $\int \frac{1}{\sqrt{t^2 - a^2}} dt = \log |t + \sqrt{t^2 - a^2}| + C$,where $t = x - \frac{3}{2}$ and $a = \frac{1}{2}$: $I = \log |(x - \frac{3}{2}) + \sqrt{(x - \frac{3}{2})^2 - (\frac{1}{2})^2}| + C$. Simplifying the expression inside the square root back to the original form: $I = \log |(x - \frac{3}{2}) + \sqrt{x^2 - 3x + 2}| + C$. Thus,the correct option is $C$.

Evaluate the integral $\int \frac{1}{\sqrt{(x-1)(x-2)}} dx$.

Similar Questions

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

For $-1 < x, y < 1$, if $\int \frac{x}{\sqrt{1+x}+\sqrt{1-x}} dx + \int \frac{y}{\sqrt{y+1}+\sqrt{y-1}} dy = A(1+x)^{3/2} + B(1-x)^{3/2} + f(y)(y+1)^{3/2} + g(y)(y-1)^{3/2} + C$, then $A f(y) + B g(y) =$

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

$\int \frac{x^5+x}{x^8+1} dx =$

$\int \frac{\sqrt{\cos 2 x}}{\sin x} d x=$

Vedclass Test Series

Exam Paper Generator

Online Exam Module