न निर्ণয় कर্রে ): \( \int e^{x} \frac{x-1}{(x+1)^{3}} d x \)

We start with the integral \[ I=\int e^{x}\frac{x-1}{(x+1)^3}\,dx. \] **Step 1. Substitute \( u = x+1 \):** Let \[ u = x+1 \quad \Rightarrow \quad x = u-1 \quad \text{and} \quad dx = du. \] Substitute into the integral. Note that \[ x-1 = (u-1)-1 = u-2 \quad \text{and} \quad (x+1)^3 = u^3. \] Also, the exponential transforms as \[ e^x = e^{u-1} = e^{-1}e^{u}. \] Thus, the integral becomes \[ I = \int e^{x}\frac{x-1}{(x+1)^3}\,dx = e^{-1} \int e^{u}\frac{u-2}{u^3}\,du. \] **Step 2. Write the integrand in terms of powers of \( u \):** Rewrite the integrand as \[ \frac{u-2}{u^3} = \frac{u}{u^3} - \frac{2}{u^3} = u^{-2} - 2u^{-3}. \] So, the integral now is \[ I = e^{-1}\left(\int e^{u}u^{-2}\,du - 2\int e^{u}u^{-3}\,du \right). \] **Step 3. Recognize the derivative of a candidate function:** Notice that the derivative of \[ \frac{e^{u}}{u^2} \] can be computed using the quotient rule or product rule: \[ \frac{d}{du}\left(\frac{e^{u}}{u^2}\right) = e^{u}u^{-2} + e^{u}\left(-2\right)u^{-3} = e^{u}\left(u^{-2}-2u^{-3}\right). \] This is exactly the integrand inside the brackets. Therefore, \[ \frac{d}{du}\left(\frac{e^{u}}{u^2}\right) = e^{u} \left(u^{-2}-2u^{-3}\right). \] **Step 4. Write the antiderivative:** Since the derivative of \(\frac{e^{u}}{u^2}\) is the integrand, we have \[ \int e^{u}\left(u^{-2}-2u^{-3}\right)\,du = \frac{e^{u}}{u^2} + C. \] Returning to the variable \( x \) (recall \( u = x+1 \)), \[ I = e^{-1}\frac{e^{u}}{u^2} + C = e^{-1}\frac{e^{x+1}}{(x+1)^2} + C = \frac{e^{x}}{(x+1)^2} + C. \] **Final Answer:** \[ \boxed{\frac{e^x}{(x+1)^2} + C} \]