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

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Aicharya · Accepted Answer

$\begin{aligned} & \text {Let } I=\int \frac{x}{\sqrt{x+1}+\sqrt{x-1}} d x \ & =\int \frac{x[\sqrt{x+1}-\sqrt{x-1}]}{x+1-x+1} d x \ & =\frac{1}{2} \int x \sqrt{x+1} d x-\frac{1}{2} \int x \sqrt{x-1} d x=\frac{1}{2} I_1-\frac{1}{2} I_2 \end{aligned}$ Now, $I_1=\int x \sqrt{x+1} d x$ Put $x+1=u \Rightarrow d x=d u$ $\begin{aligned} \therefore I_1 & =\int(u-1) \sqrt{u} d x=\int\left(u^{\frac{3}{2}}-u^{\frac{1}{2}}\right) d x \ & =\frac{2}{5} u^{\frac{5}{2}}-\frac{2}{3} u^{\frac{3}{2}}+c_1=\frac{2}{5}(x+1)^{\frac{5}{2}}-\frac{2}{3}(x+1)^{\frac{3}{2}}+c_1 \ & =2(x+1)^{\frac{3}{2}}\left[\frac{1}{5}(x+1)-\frac{1}{3}\right]+c_1 \ & =2(x+1)^{\frac{3}{2}}\left[\frac{3 x+3-5}{15}\right]+c_1 \ & =\frac{2(3 x-2)}{15}(x+1)^{\frac{3}{2}}+c_1 \end{aligned}$ Again, $I_2=\int x \sqrt{x-1} d x$ Put $x-1=v$ $\begin{aligned} \Rightarrow \quad & d x=d v \ \therefore \quad & I_2=\int(v+1) \sqrt{v} d v=\int\left(v^{\frac{3}{2}}+v^{\frac{1}{2}}\right) d v \ = & \frac{2}{5} v^{\frac{5}{2}}+\frac{2}{3} v^{\frac{3}{2}}+c_2=\frac{2}{5}(x-1)^{\frac{5}{2}}+\frac{2}{3}(x-1)^{\frac{3}{2}}+c_2 \ = & 2(x-1)^{\frac{3}{2}}\left[\frac{1}{5}(x-1)+\frac{1}{3}\right]+c_2 \ = & 2(x-1)^{\frac{3}{2}}\left[\frac{3 x-3+5}{15}\right]+c_2 \ = & \frac{2(3 x+2)}{15}(x-1)^{\frac{3}{2}}+c_2 \ \therefore I= & \frac{1}{2} I_1-\frac{1}{2} I_2 \end{aligned}$ $\begin{gathered} =\frac{(3 x-2)}{15}(x+1)^{\frac{3}{2}}-\frac{(3 x+2)}{15}(x-1)^{\frac{3}{2}}+c \ \quad\left[\text {where } c=c_1+c_2\right] \ \therefore A(x)=\frac{1}{15}(3 x-2) \text { and } B(x)=\frac{-1}{15}(3 x+2] \ \therefore A(x)+B(x)=\frac{1}{15}(3 x-2)-\frac{1}{15}(3 x+2) \ =\frac{-4}{15} \end{gathered}$

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

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