int frac{d x}{(x+100) sqrt{x+99}}=f(x)+c Righ | vedclass

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Question

(C) Let $I = \int \frac{d x}{(x+100) \sqrt{x+99}}$.We can rewrite the denominator as $(x+99)+1$,so $I = \int \frac{d x}{((\sqrt{x+99})^2+1) \sqrt{x+99

Vedclass · Accepted Answer

$2 	an ^{-1}(\sqrt{x+99})$

Vedclass · Answer

(C) Let $I = \int \frac{d x}{(x+100) \sqrt{x+99}}$.We can rewrite the denominator as $(x+99)+1$,so $I = \int \frac{d x}{((\sqrt{x+99})^2+1) \sqrt{x+99}}$.Substitute $t = \sqrt{x+99}$,then $t^2 = x+99$,which implies $2t \, dt = dx$.Substituting these into the integral:$I = \int \frac{2t \, dt}{(t^2+1)t} = \int \frac{2 \, dt}{t^2+1}$.Integrating,we get $I = 2 	an^{-1}(t) + c$.Substituting back $t = \sqrt{x+99}$,we get $I = 2 	an^{-1}(\sqrt{x+99}) + c$.Comparing this with $f(x) + c$,we find $f(x) = 2 	an^{-1}(\sqrt{x+99})$.

$\int \frac{d x}{(x+100) \sqrt{x+99}}=f(x)+c \Rightarrow f(x)$

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