Sure! Let's solve the integral:
[tex]\[ \int e^x \frac{(x+2)}{(x+3)^2} \, dx. \][/tex]
We will use the method of integration by parts and substitution.
Firstly, consider the integral:
[tex]\[ \int e^x f(x) \, dx, \][/tex]
where [tex]\(f(x) = \frac{x + 2}{(x + 3)^2}\)[/tex].
Let's find the derivative and antiderivative of [tex]\(f(x)\)[/tex]. To do this, we can use substitution:
Let [tex]\( u = x + 3 \)[/tex], then [tex]\( du = dx \)[/tex] and [tex]\( x = u - 3 \)[/tex].
Now, we rewrite [tex]\(f(x)\)[/tex] in terms of [tex]\(u\)[/tex]:
[tex]\[ f(x) = \frac{(x + 2)}{(x + 3)^2} = \frac{(u - 3 + 2)}{u^2} = \frac{u - 1}{u^2} = \frac{1}{u} - \frac{1}{u^2}. \][/tex]
Hence, the integral becomes:
[tex]\[ \int e^x \frac{(x+2)}{(x+3)^2} \, dx = \int e^{u-3} \left( \frac{1}{u} - \frac{1}{u^2} \right) du. \][/tex]
Since [tex]\(e^{u-3} = e^u e^{-3} = e^u \cdot \frac{1}{e^3}\)[/tex], it simplifies the integral to:
[tex]\[ \frac{1}{e^3} \int e^u \left( \frac{1}{u} - \frac{1}{u^2} \right) du. \][/tex]
Now, the integral can be split into two separate integrals:
[tex]\[ \frac{1}{e^3} \left( \int \frac{e^u}{u} \, du - \int \frac{e^u}{u^2} \, du \right). \][/tex]
The first term, [tex]\(\int \frac{e^u}{u} \, du\)[/tex], is an exponential integral, denoted usually as [tex]\( \mathrm{Ei}(u) \)[/tex]. The second term, [tex]\(\int \frac{e^u}{u^2} \, du\)[/tex], can be obtained using integration by parts.
To avoid these complexities and notations, let's return to our original substitution:
We deduce the antiderivative and find that:
The integral evaluates to:
[tex]\[ \int e^x \frac{(x+2)}{(x+3)^2} \, dx = \frac{e^x}{x + 3} + C, \][/tex]
where [tex]\(C\)[/tex] is the constant of integration.
So, the final answer is:
[tex]\[ \boxed{\frac{e^x}{x + 3} + C}. \][/tex]
