int frac{d x}{(x+1) sqrt{4 x+3}} is equal to | vedclass

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(C) Let $I = \int \frac{d x}{(x+1) \sqrt{4 x+3}}$.Substitute $4x + 3 = t^2$, which implies $4dx = 2tdt$ or $dx = \frac{1}{2}tdt$.Also, $x = \frac{t^2

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$2 	an ^{-1} \sqrt{4 x+3}+c$

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(C) Let $I = \int \frac{d x}{(x+1) \sqrt{4 x+3}}$.Substitute $4x + 3 = t^2$, which implies $4dx = 2tdt$ or $dx = \frac{1}{2}tdt$.Also, $x = \frac{t^2 - 3}{4}$, so $x + 1 = \frac{t^2 - 3}{4} + 1 = \frac{t^2 + 1}{4}$.Substituting these into the integral:$I = \int \frac{\frac{1}{2} t dt}{(\frac{t^2 + 1}{4}) t} = \int \frac{\frac{1}{2} dt}{\frac{t^2 + 1}{4}} = 2 \int \frac{dt}{t^2 + 1}$.Integrating, we get $I = 2 	an^{-1}(t) + c$.Substituting back $t = \sqrt{4x + 3}$, we get $I = 2 	an^{-1}(\sqrt{4x + 3}) + c$.

$\int \frac{d x}{(x+1) \sqrt{4 x+3}}$ is equal to

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