The value of int {frac{1}{{{{(x - 5)}^2}}},dx | vedclass

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Question

(B) Let $I = \int {\frac{1}{{{{(x - 5)}^2}}}dx} $.We can rewrite the integrand as $(x - 5)^{-2}$.Using the power rule for integration,$\int {x^n dx} =

Vedclass · Accepted Answer

$ - \frac{1}{{x - 5}} + c$

Vedclass · Answer

(B) Let $I = \int {\frac{1}{{{{(x - 5)}^2}}}dx} $.We can rewrite the integrand as $(x - 5)^{-2}$.Using the power rule for integration,$\int {x^n dx} = \frac{x^{n+1}}{n+1} + c$ (where $n 
eq -1$):$I = \int {(x - 5)^{-2} dx} = \frac{{{{(x - 5)}^{ - 2 + 1}}}}{{ - 2 + 1}} + c$$I = \frac{{{{(x - 5)}^{ - 1}}}}{{ - 1}} + c$$I = - \frac{1}{{(x - 5)}} + c$.

The value of $\int {\frac{1}{{{{(x - 5)}^2}}}\,dx} $ is

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