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

Question

UpStudy AI · Accepted Answer

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:**

$$
\boxed{\frac{e^x}{(x+1)^2} + C}
$$
The integral simplifies to $$ \frac{e^{x}}{(x+1)^2} + C $$.

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