Evaluate the integral: int frac{x^2 , dx}{(x^ | vedclass

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Question

(D) Let $t = x^2$. Then the integrand becomes $\frac{t}{(t+2)(t+5)}$.Using partial fractions: $\frac{t}{(t+2)(t+5)} = \frac{A}{t+2} + \frac{B}{t+5}$.$

Vedclass · Accepted Answer

$-\frac{\sqrt{2}}{3} 	an^{-1} \left( \frac{x}{\sqrt{2}} ight) + \frac{\sqrt{5}}{3} 	an^{-1} \left( \frac{x}{\sqrt{5}} ight) + c$

Vedclass · Answer

(D) Let $t = x^2$. Then the integrand becomes $\frac{t}{(t+2)(t+5)}$.Using partial fractions: $\frac{t}{(t+2)(t+5)} = \frac{A}{t+2} + \frac{B}{t+5}$.$t = A(t+5) + B(t+2)$.For $t = -2$, $-2 = A(3) \implies A = -2/3$.For $t = -5$, $-5 = B(-3) \implies B = 5/3$.So, $\int \frac{x^2 \, dx}{(x^2 + 2)(x^2 + 5)} = \int \left( \frac{-2/3}{x^2 + 2} + \frac{5/3}{x^2 + 5} ight) dx$.$= -\frac{2}{3} \int \frac{dx}{x^2 + (\sqrt{2})^2} + \frac{5}{3} \int \frac{dx}{x^2 + (\sqrt{5})^2}$.Using $\int \frac{dx}{x^2 + a^2} = \frac{1}{a} 	an^{-1} \frac{x}{a} + c$:$= -\frac{2}{3} \cdot \frac{1}{\sqrt{2}} 	an^{-1} \frac{x}{\sqrt{2}} + \frac{5}{3} \cdot \frac{1}{\sqrt{5}} 	an^{-1} \frac{x}{\sqrt{5}} + c$.$= -\frac{\sqrt{2}}{3} 	an^{-1} \frac{x}{\sqrt{2}} + \frac{\sqrt{5}}{3} 	an^{-1} \frac{x}{\sqrt{5}} + c$.

Evaluate the integral: $\int \frac{x^2 \, dx}{(x^2 + 2)(x^2 + 5)}$

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