The primitive of (x^2 + 4)^{-1/2} with respec | vedclass

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Question

(C) Let $u = x^2 + 2$. Then $du = 2x \, dx$. We want to find the integral $\int (x^2 + 4)^{-1/2} \, d(x^2 + 2)$. Since $d(x^2 + 2) = 2x \, dx$,the int

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$2\sqrt{x^2 + 4} + C$

Vedclass · Answer

(C) Let $u = x^2 + 2$. Then $du = 2x \, dx$. We want to find the integral $\int (x^2 + 4)^{-1/2} \, d(x^2 + 2)$. Since $d(x^2 + 2) = 2x \, dx$,the integral becomes $\int \frac{1}{\sqrt{x^2 + 4}} \cdot 2x \, dx$. Let $t = x^2 + 4$. Then $dt = 2x \, dx$. The integral becomes $\int \frac{1}{\sqrt{t}} \, dt = \int t^{-1/2} \, dt$. Using the power rule for integration,$\int t^n \, dt = \frac{t^{n+1}}{n+1} + C$,we get: $\frac{t^{1/2}}{1/2} + C = 2\sqrt{t} + C$. Substituting back $t = x^2 + 4$,we get $2\sqrt{x^2 + 4} + C$.

The primitive of $(x^2 + 4)^{-1/2}$ with respect to $x^2 + 2$ is equal to :-

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