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 graph of an exponential function $y = 2 + ae^{-x}$ is shown in the figure. What is the value of $a$?

If $x^3+3xy+y^3=1$,then the correct option$(s)$ is/are:-
$(A) \left(\frac{dy}{dx}\right)_{(1,1)}=-1$
$(B) \left(\frac{dy}{dx}\right)_{(1,1)}=-2$
$(C) \left(\frac{dy}{dx}\right)_{(1,0)}=-1$
$(D) \left(\frac{dy}{dx}\right)_{(1,0)}=-3$

What is the value of $\frac{1}{2}$?

The value of $1+\frac{1}{4}+\frac{1}{16}+\frac{1}{64}+\ldots$ up to $\infty$ is

If $F = \frac{2}{\sin \theta + \sqrt{3} \cos \theta}$,then the minimum value of $F$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module