int frac{3}{(2-x)^2} , dx is equal to :- | vedclass

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

Question

(A) To evaluate the integral $\int \frac{3}{(2-x)^2} \, dx$,we use the substitution method.Let $t = 2-x$.Then,$dt = -dx$,which implies $dx = -dt$.Subs

Vedclass · Accepted Answer

$\frac{3}{(2-x)}+C$

Vedclass · Answer

(A) To evaluate the integral $\int \frac{3}{(2-x)^2} \, dx$,we use the substitution method.Let $t = 2-x$.Then,$dt = -dx$,which implies $dx = -dt$.Substituting these into the integral:$\int \frac{3}{t^2} (-dt) = -3 \int t^{-2} \, dt$.Using the power rule $\int t^n \, dt = \frac{t^{n+1}}{n+1} + C$:$-3 \left( \frac{t^{-2+1}}{-2+1} \right) + C = -3 \left( \frac{t^{-1}}{-1} \right) + C$.$= 3t^{-1} + C = \frac{3}{t} + C$.Substituting back $t = 2-x$,we get $\frac{3}{2-x} + C$.

$\int \frac{3}{(2-x)^2} \, dx$ is equal to $:-$

Similar Questions

The value of $\frac{d}{dx} (\cos 4x^2)$ is .......

$\int \frac{dx}{{(2ax - x^2)}^{1/2}} = a^n \sin^{-1} \left( \frac{x}{a} - 1 \right)$. In this formula,$n =$ . . . . . . .

The side of a square is increasing at the rate of $0.2\,cm/s$. The rate of increase of perimeter with respect to time is $...........\,cm/s$.

The value of the definite integral $\int_{0}^{\pi/4} \sin(2x) \, dx$ is:

Which of the following relation$(s)$ is/are correct?
$(A)\ \sin (90^{\circ}+\theta)=\cos (-\theta)$
$(B)\ \sin (180^{\circ}-\theta)=\cos (90^{\circ}-\theta)$
$(C)\ \sin (360^{\circ}-\theta)=\cos (360^{\circ}-\theta)$
$(D)\ \sin (180^{\circ}+\theta)=\cos (90^{\circ}-\theta)$

Vedclass Test Series

Exam Paper Generator

Online Exam Module