$I=\int_1^m\ \frac{e^x(x^2+1)}{(x+1)^2}dx\ হলে,\left(\frac{(m+1)}{(m-1)}\right)^3\ I^3এর\ মান\ নির্ণয়\ কর।$

$\begin{array}{l}I=\int_{1}^{m} \frac{e^{x}\left(x^{2}+1\right)}{(x+1)^{2}} d x \\ \Rightarrow I=\int_{1}^{m} e^{x}\left\{\frac{x^{2}}{(x+1)^{2}}+\frac{1}{(x+1)^{2}}\right\} d x\end{array}$

$\begin{array}{l}\Rightarrow I=\int_{1}^{m} e^{x}\left\{\frac{x^{2}-1+1}{(x+1)^{2}}+\frac{1}{(x+1)^{2}} \int d x\right. \\ \Rightarrow I=\int_{1}^{m} e^{x}\left\{\frac{x^{2}-1}{(x+1)^{2}}+\frac{1}{(x+1)^{2}}+\frac{1}{(x+1)^{2}}\right\} d x\end{array}$

$\Rightarrow I=\int_{1}^{m} e^{x}\left\{\frac{(x+1)(x-1)}{(x+1)(x+1)}+\frac{2}{(x+1)^{2}}\right\} d x$

$\Rightarrow I=\int_{1}^{m} e^{x}\left\{\frac{x-1}{x+1}+\frac{2}{(x+1)^{2}}\right\} d x$

We know,

$\begin{array}{l} \int e^{x}\left\{f(x)+f^{\prime}(x)\right\} d x \\ =e^{x} f(x)+c . \end{array}$

Let,

$\begin{aligned} & f(x)=\frac{x-1}{x+1} \\ \Rightarrow & f^{\prime}(x)=\frac{(x+1) \frac{d}{d x}(x-1)-(x-1) \frac{d}{d x}(x+1)}{(x+1)^{2}} \end{aligned}$

$\begin{array}{l} \Rightarrow f^{\prime}(x)=\frac{ x+1-x+1}{(x+1)^{2}} \\ \Rightarrow f^{\prime}(x)=\frac{2}{(x+1)^{2}}\end{array}$

$\begin{array}{l}\therefore I=\int_{1}^{m} e^{x}\left\{\frac{x-1}{x+1}+\frac{2}{(x+1) ²}\right\} d x \\ \Rightarrow I=\left[e^{x}\left(\frac{x-1}{x+1}\right)\right]_{1}^{m}\end{array}$

$\begin{array}{l}\Rightarrow I=e^{m}\left(\frac{m-1}{m+1}\right)-e\left(\frac{1-1}{1+1}\right) \\ \Rightarrow I=e^{m}\left(\frac{m-1}{m+1}\right)\end{array}$

$\begin{array}{l}\Rightarrow\left(\frac{m+1}{m-1}\right) I=e^{m} \\ \Rightarrow\left(\frac{m+1}{m-1}\right)^{3} I^{3}=e^{3 m}\end{array}$