Let’s solve each integral step by step:

1.

$\int_{2}^{\infty} \frac{(x+3)}{(x-1)(x^2+1)} \, dx$

Steps:

1. Decompose the fraction using partial fractions: $\frac{x+3}{(x-1)(x^2+1)} = \frac{A}{x-1} + \frac{Bx+C}{x^2+1}$ Solve for $A$, $B$, and $C$.

2. Integrate each part:

   • The term $\frac{1}{x-1}$ leads to a logarithm.
   • The term $\frac{Bx+C}{x^2+1}$ splits further into arctangent and logarithmic terms.

3. Apply limits ($x=2$ to $\infty$) carefully.

Let me know if you'd like detailed solutions for partial fraction decomposition and evaluation.

2.

$\int_{-\infty}^\infty e^x \, dx$

The function $e^x$ diverges as $x \to \infty$. Therefore, the integral does not converge.

3.

$\int_{0}^\infty x^2 e^{-x} \, dx$

Steps:

1. This integral is solved using the Gamma function. Specifically: $\int_{0}^\infty x^n e^{-x} \, dx = \Gamma(n+1),$ where $\Gamma(n+1) = n!$.

2. Here, $n = 2$: $\int_{0}^\infty x^2 e^{-x} \, dx = \Gamma(3) = 2! = 2.$

4.

$\int_{1}^\infty \frac{e^{-\sqrt{x}}}{\sqrt{x}} \, dx$

Substitution:

Let $u = \sqrt{x}$, so $x = u^2$ and $dx = 2u \, du$. The limits transform as follows:

• When $x = 1$, $u = 1$.
• When $x \to \infty$, $u \to \infty$.

The integral becomes: $\int_{1}^\infty \frac{e^{-\sqrt{x}}}{\sqrt{x}} \, dx = \int_{1}^\infty e^{-u} \cdot 2 \, du = 2 \int_{1}^\infty e^{-u} \, du.$

Solve:

[ 2 \int_{1}^\infty e^{-u} , du = 2 \left[-e^{-u}\right]_1^\infty = 2 \left[0 - (-e^{-1})\right] = 2e^{-1} = \frac{2}{e}. ]

Let me know which part you'd like to explore in further detail!

5 Related Questions:

1. How do you perform partial fraction decomposition for complex denominators?
2. What are the convergence conditions for improper integrals?
3. How is the Gamma function used to solve integrals?
4. What substitutions simplify integrals with roots in the exponent?
5. How can improper integrals with infinity be handled systematically?

Tip:

For improper integrals, always check for convergence using limits before attempting evaluation.