The value of ∫(x-4)/(x^2√(x-2))dx is equal to (where, C is the constant of integration)

Correct answer: (B) -2/x√(x-2)+C. ∫(x-4)/(x^2√(x-2))dx ∫1/(x^2)-4/(x^3)√(1x-2x^2)dx Let 1/x-2/(x^2)=t -1x^2+4x^3dx=dt ∫(-dt√t=-2√t+C=-2√(1x-2x^2)+C=-2)/x√(x-2)+C

Q28 / 104 · NTA Abhyas (JEE Main) 2020 · Mathematics · Indefinite Integration

The value of $\int \frac{(x - 4)}{x^{2} \sqrt{x - 2}} d x$ is equal to (where, $C$ is the constant of integration)

A
$2 x \sqrt{x - 2} + C$
Correct answer
B
$- \frac{2}{x} \sqrt{x - 2} + C$
C
$\frac{\sqrt{x - 2}}{x} + C$
D
$\frac{x}{\sqrt{x - 2}} + C$

Solution

\int \frac{x - 4}{x^{2} \sqrt{x - 2}} d x

\int \frac{(\frac{1}{x^{2}} - \frac{4}{x^{3}})}{\sqrt{\frac{1}{x} - \frac{2}{x^{2}}}} d x

Let

\frac{1}{x} - \frac{2}{x^{2}} = t

(- \frac{1}{x^{2}} + \frac{4}{x^{3}}) d x = d t

\int \frac{- d t}{\sqrt{t}} = - 2 \sqrt{t} + C

= - 2 \sqrt{\frac{1}{x} - \frac{2}{x^{2}}} + C

= - \frac{2}{x} \sqrt{x - 2} + C

Hence, the correct answer is option: 'B'.

The value of ∫x-4x2x-2dx is equal to (where, C is the constant of integration)

Solution

More Indefinite Integration Questions

The value of $\int \frac{(x - 4)}{x^{2} \sqrt{x - 2}} d x$ is equal to (where, $C$ is the constant of integration)