int {frac{{{x^2}}}{{{x^2} + 4}},dx} is equal | vedclass

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(A) Let $I = \int \frac{x^2}{x^2 + 4} dx$. We can rewrite the integrand as: $I = \int \frac{x^2 + 4 - 4}{x^2 + 4} dx$ $I = \int \left( 1 - \frac{4}{x^

Vedclass · Accepted Answer

$x - 2	an^{-1}(x/2) + c$

Vedclass · Answer

(A) Let $I = \int \frac{x^2}{x^2 + 4} dx$. We can rewrite the integrand as: $I = \int \frac{x^2 + 4 - 4}{x^2 + 4} dx$ $I = \int \left( 1 - \frac{4}{x^2 + 4} ight) dx$ $I = \int 1 dx - 4 \int \frac{1}{x^2 + 2^2} dx$ Using the standard integral formula $\int \frac{1}{x^2 + a^2} dx = \frac{1}{a} 	an^{-1}(\frac{x}{a}) + c$: $I = x - 4 \left( \frac{1}{2} 	an^{-1}(\frac{x}{2}) ight) + c$ $I = x - 2 	an^{-1}(\frac{x}{2}) + c$.

$\int {\frac{{{x^2}}}{{{x^2} + 4}}\,dx} $ is equal to

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