If intlimits_0^2 375 x^5 (1 + x^2)^{-4} dx = | vedclass

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(B) Let $I = \int_0^2 \frac{375 x^5}{(1 + x^2)^4} dx$. Substitute $u = 1 + x^2$,then $du = 2x dx$,so $x^2 = u - 1$ and $x dx = \frac{du}{2}$. When $x

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$5$

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(B) Let $I = \int_0^2 \frac{375 x^5}{(1 + x^2)^4} dx$. Substitute $u = 1 + x^2$,then $du = 2x dx$,so $x^2 = u - 1$ and $x dx = \frac{du}{2}$. When $x = 0$,$u = 1$. When $x = 2$,$u = 5$. $I = \int_1^5 \frac{375 (x^2)^2 x}{(1 + x^2)^4} dx = \int_1^5 \frac{375 (u - 1)^2}{u^4} \cdot \frac{du}{2}$. $I = \frac{375}{2} \int_1^5 \frac{u^2 - 2u + 1}{u^4} du = \frac{375}{2} \int_1^5 (u^{-2} - 2u^{-3} + u^{-4}) du$. $I = \frac{375}{2} \left[ -u^{-1} + u^{-2} - \frac{1}{3} u^{-3} \right]_1^5$. $I = \frac{375}{2} \left[ (-\frac{1}{5} + \frac{1}{25} - \frac{1}{375}) - (-1 + 1 - \frac{1}{3}) \right]$. $I = \frac{375}{2} \left[ \frac{-75 + 15 - 1}{375} + \frac{1}{3} \right] = \frac{375}{2} \left[ \frac{-61}{375} + \frac{125}{375} \right] = \frac{375}{2} \cdot \frac{64}{375} = \frac{64}{2} = 32$. Since $32 = 2^5$,we have $2^n = 2^5$,which implies $n = 5$.

If $\int\limits_0^2 375 x^5 (1 + x^2)^{-4} dx = 2^n$,then the value of $n$ is:

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