If int frac{x}{sqrt{x+1}+sqrt{x-1}} d x=A(x)( | vedclass

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(B) Let $I=\int \frac{x}{\sqrt{x+1}+\sqrt{x-1}} d x$. Rationalizing the denominator,we get: $I=\int \frac{x(\sqrt{x+1}-\sqrt{x-1})}{(x+1)-(x-1)} d x =

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$-\frac{4}{15}$

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(B) Let $I=\int \frac{x}{\sqrt{x+1}+\sqrt{x-1}} d x$. Rationalizing the denominator,we get: $I=\int \frac{x(\sqrt{x+1}-\sqrt{x-1})}{(x+1)-(x-1)} d x = \frac{1}{2} \int x \sqrt{x+1} d x - \frac{1}{2} \int x \sqrt{x-1} d x = \frac{1}{2} I_1 - \frac{1}{2} I_2$. For $I_1 = \int x \sqrt{x+1} d x$,let $u = x+1$,then $x = u-1$ and $dx = du$: $I_1 = \int (u-1) \sqrt{u} du = \int (u^{3/2} - u^{1/2}) du = \frac{2}{5} u^{5/2} - \frac{2}{3} u^{3/2} + c_1 = 2(x+1)^{3/2} [\frac{1}{5}(x+1) - \frac{1}{3}] + c_1 = \frac{2(3x-2)}{15}(x+1)^{3/2} + c_1$. For $I_2 = \int x \sqrt{x-1} d x$,let $v = x-1$,then $x = v+1$ and $dx = dv$: $I_2 = \int (v+1) \sqrt{v} dv = \int (v^{3/2} + v^{1/2}) dv = \frac{2}{5} v^{5/2} + \frac{2}{3} v^{3/2} + c_2 = 2(x-1)^{3/2} [\frac{1}{5}(x-1) + \frac{1}{3}] + c_2 = \frac{2(3x+2)}{15}(x-1)^{3/2} + c_2$. Substituting back into $I$: $I = \frac{1}{2} [\frac{2(3x-2)}{15}(x+1)^{3/2}] - \frac{1}{2} [\frac{2(3x+2)}{15}(x-1)^{3/2}] + C = \frac{3x-2}{15}(x+1)^{3/2} - \frac{3x+2}{15}(x-1)^{3/2} + C$. Comparing with $A(x)(x+1)^{3/2} + B(x)(x-1)^{3/2} + C$,we get $A(x) = \frac{3x-2}{15}$ and $B(x) = -\frac{3x+2}{15}$. Thus,$A(x) + B(x) = \frac{3x-2 - 3x - 2}{15} = -\frac{4}{15}$.

If $\int \frac{x}{\sqrt{x+1}+\sqrt{x-1}} d x=A(x)(x+1)^{\frac{3}{2}}+B(x)(x-1)^{\frac{3}{2}}+C$,then $A(x)+B(x)=$

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