Step 1a

We can rewrite the integrand as:

\$ (2x + 1) / x^2 = 2x/x^2 + 1/x^2 = 2/x + 1/x^2 \$

Splitting the integral, we get:

\$ \int 2/x dx + \int 1/x^2 dx \$

Explanation: Here, we rewrote and split the integrand.

Step 1b

The integration formulas are:

\$ \int 1/x dx = ln| x | + c , \int x^n dx = (x^(n+1)) /(n+1) + c \$

Now applying the formulas we get:

\$ 2 \int (1/x) dx + \int (x^(-2) ) dx \$

After integration, we get:

\$ 2 ln| x | + (x^(-2 + 1))/(-2+1) + c \$

After simplification :

\$ 2ln| x | + (x^(-1))/(-1) + c \$

\$ 2ln| x | - 1/x + c \$

Explanation: Here, after applied the formulas then we get \$ 2ln| x | - 1/x + c \$

Step 2a

The given series is:

\$ \sum_{n=1}^\infty (2^(n-1) / n) \$

Simplify \$2^(n-1)\$ = \$2^n 2^(-1)\$

⇒ \$\sum_{n=1}^\infty (2^n * 2^-1) / n\$

Apply the constant multiplication rule: \$ ∑ c a_n = c ∑a_n\$

⇒ \$ 2^(-1) \sum_{n=1}^\infty 2^n / (n) \$

Apply exponent rule : \$a^(-1) = 1/a\$

⇒ \$1/2 \sum_{n=1}^\infty 2^n / (n)\$

Apply series ratio test: diverges

⇒ \$1/2\$ diverges

Explanation: First, simplify the given series, then apply the constant rule. After simplification, apply the exponent, and finally get the answer. Given series \$1/2\$ diverges.

Note: Series ratio test

Step 2b

To find the sum of the series, we can use the formula for the sum of an infinite geometric series:

\$ S = a / (1 - r) \$

where a is the first term and r is the common ratio. In this case, a = 1 and \$ r = 1/2\$. Thus,

\$ S = 1 / (1 - 1/2) = 2 \$

Therefore, the sum of the given series is 2.

Explanation: After replacing r in the sum of the series formula, the sum of the given series simplifies to 2.