int frac{d x}{sqrt{left(5+2 x+x^2right)^3}} i | vedclass

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Question

(D) We have $5+2 x+x^2 = (x+1)^2 + 4$. Let $x+1 = z$, then $dx = dz$. Substituting this into the integral, we get $I = \int \frac{dz}{(z^2+4)^{3/2}}$.

Vedclass · Accepted Answer

$\frac{1}{4} \frac{x+1}{\sqrt{5+2 x+x^2}}+C$

Vedclass · Answer

(D) We have $5+2 x+x^2 = (x+1)^2 + 4$. Let $x+1 = z$, then $dx = dz$. Substituting this into the integral, we get $I = \int \frac{dz}{(z^2+4)^{3/2}}$. Let $z = 2 	an 	heta$, then $dz = 2 \sec^2 	heta \ d	heta$. Also, $(z^2+4)^{3/2} = (4 	an^2 	heta + 4)^{3/2} = (4 \sec^2 	heta)^{3/2} = 8 \sec^3 	heta$. Therefore, $I = \int \frac{2 \sec^2 	heta \ d	heta}{8 \sec^3 	heta} = \frac{1}{4} \int \cos 	heta \ d	heta = \frac{1}{4} \sin 	heta + C$. Since $	an 	heta = \frac{z}{2}$, we have $\sin 	heta = \frac{z}{\sqrt{z^2+4}}$. Thus, $I = \frac{1}{4} \cdot \frac{z}{\sqrt{z^2+4}} + C = \frac{x+1}{4 \sqrt{(x+1)^2+4}} + C = \frac{x+1}{4 \sqrt{5+2x+x^2}} + C$.

$\int \frac{d x}{\sqrt{\left(5+2 x+x^2\right)^3}}$ is equal to

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