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

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

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

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

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

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