int frac{d x}{(x+2) sqrt{x+1}} = | vedclass

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(B) Let $I = \int \frac{d x}{(x+2) \sqrt{x+1}}$.Substitute $\sqrt{x+1} = t$,which implies $x+1 = t^2$,so $x = t^2 - 1$ and $dx = 2t \, dt$.Substitutin

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

Vedclass · Answer

(B) Let $I = \int \frac{d x}{(x+2) \sqrt{x+1}}$.Substitute $\sqrt{x+1} = t$,which implies $x+1 = t^2$,so $x = t^2 - 1$ and $dx = 2t \, dt$.Substituting these into the integral:$I = \int \frac{2t \, dt}{(t^2 - 1 + 2) \cdot t}$$I = \int \frac{2t \, dt}{(t^2 + 1) \cdot t}$$I = 2 \int \frac{dt}{t^2 + 1}$Using the standard integral formula $\int \frac{dt}{t^2 + 1} = 	an^{-1}(t) + c$:$I = 2 	an^{-1}(t) + c$Substituting back $t = \sqrt{x+1}$:$I = 2 	an^{-1}(\sqrt{x+1}) + c$.

$\int \frac{d x}{(x+2) \sqrt{x+1}} = $

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