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

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(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

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

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 $:-$

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