If f(x)+k is obtained by evaluating int frac{ | vedclass

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(A) Let $I = \int \frac{x^3}{(1+x^2)^3} dx$.Method $1$: Using $x = 	an 	heta$,$dx = \sec^2 	heta d	heta$.$I = \int \frac{	an^3 	heta \sec^2 	he

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$\frac{1}{4}$

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(A) Let $I = \int \frac{x^3}{(1+x^2)^3} dx$.Method $1$: Using $x = 	an 	heta$,$dx = \sec^2 	heta d	heta$.$I = \int \frac{	an^3 	heta \sec^2 	heta}{(1+	an^2 	heta)^3} d	heta = \int \frac{	an^3 	heta \sec^2 	heta}{\sec^6 	heta} d	heta = \int 	an^3 	heta \cos^4 	heta d	heta = \int \sin^3 	heta \cos 	heta d	heta$.Let $\sin 	heta = p$,then $\cos 	heta d	heta = dp$.$I = \int p^3 dp = \frac{p^4}{4} + k = \frac{\sin^4 	heta}{4} + k$.Since $	an 	heta = x$,$\sin 	heta = \frac{x}{\sqrt{1+x^2}}$,so $I = \frac{x^4}{4(1+x^2)^2} + k$. Thus $f(x) = \frac{x^4}{4(1+x^2)^2}$.Method $2$: Using $1+x^2 = z$,$2x dx = dz \Rightarrow x dx = \frac{1}{2} dz$.$I = \int \frac{x^2 \cdot x dx}{(1+x^2)^3} = \int \frac{(z-1) \cdot \frac{1}{2} dz}{z^3} = \frac{1}{2} \int (z^{-2} - z^{-3}) dz = \frac{1}{2} [-\frac{1}{z} + \frac{1}{2z^2}] + c = -\frac{1}{2(1+x^2)} + \frac{1}{4(1+x^2)^2} + c$.Thus $g(x) = -\frac{1}{2(1+x^2)} + \frac{1}{4(1+x^2)^2}$.Now,$f(x) - g(x) = \frac{x^4}{4(1+x^2)^2} + \frac{1}{2(1+x^2)} - \frac{1}{4(1+x^2)^2} = \frac{x^4 - 1 + 2(1+x^2)}{4(1+x^2)^2} = \frac{x^4 + 2x^2 + 1}{4(1+x^2)^2} = \frac{(x^2+1)^2}{4(1+x^2)^2} = \frac{1}{4}$.Therefore,$f(x) - g(x) + k - c = \frac{1}{4} + k - c$,which is a constant.

If $f(x)+k$ is obtained by evaluating $\int \frac{x^3}{\left(1+x^2\right)^3} d x$ using the substitution $x=\tan \theta$,and $g(x)+c$ is obtained by evaluating $\int \frac{x^3}{\left(1+x^2\right)^3} d x$ using the substitution $x^2+1=z$,then $f(x)-g(x)+k-c=$

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