int frac{d x}{(x+1) sqrt{x^2+4}} = | vedclass

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Question

(A) Let $I = \int \frac{d x}{(x+1) \sqrt{x^2+4}}$.Substitute $x+1 = \frac{1}{t}$, then $x = \frac{1}{t} - 1 = \frac{1-t}{t}$ and $dx = -\frac{1}{t^2}

Vedclass · Accepted Answer

$-\frac{1}{\sqrt{5}} \sinh ^{-1}\left(\frac{4-x}{2(x+1)}\right)+c$

Vedclass · Answer

(A) Let $I = \int \frac{d x}{(x+1) \sqrt{x^2+4}}$.Substitute $x+1 = \frac{1}{t}$, then $x = \frac{1}{t} - 1 = \frac{1-t}{t}$ and $dx = -\frac{1}{t^2} dt$.Substituting these into the integral:$I = \int \frac{-\frac{1}{t^2} dt}{\frac{1}{t} \sqrt{(\frac{1-t}{t})^2 + 4}} = -\int \frac{dt}{t \sqrt{\frac{1-2t+t^2+4t^2}{t^2}}} = -\int \frac{dt}{\sqrt{5t^2-2t+1}}$.Completing the square in the denominator:$5t^2 - 2t + 1 = 5(t^2 - \frac{2}{5}t + \frac{1}{5}) = 5((t-\frac{1}{5})^2 + \frac{4}{25}) = 5(t-\frac{1}{5})^2 + \frac{4}{5}$.$I = -\int \frac{dt}{\sqrt{5(t-\frac{1}{5})^2 + \frac{4}{5}}} = -\frac{1}{\sqrt{5}} \int \frac{dt}{\sqrt{(t-\frac{1}{5})^2 + (\frac{2}{5})^2}}$.Using the formula $\int \frac{du}{\sqrt{u^2+a^2}} = \sinh^{-1}(\frac{u}{a})$:$I = -\frac{1}{\sqrt{5}} \sinh^{-1}(\frac{t-\frac{1}{5}}{2/5}) + C = -\frac{1}{\sqrt{5}} \sinh^{-1}(\frac{5t-1}{2}) + C$.Substituting $t = \frac{1}{x+1}$ back:$I = -\frac{1}{\sqrt{5}} \sinh^{-1}(\frac{\frac{5}{x+1}-1}{2}) + C = -\frac{1}{\sqrt{5}} \sinh^{-1}(\frac{5-x-1}{2(x+1)}) + C = -\frac{1}{\sqrt{5}} \sinh^{-1}(\frac{4-x}{2(x+1)}) + C$.

$\int \frac{d x}{(x+1) \sqrt{x^2+4}} = $

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