Exponential And Logarithmic Functions

Let .
Let .
Then .
Use partial fractions for .
Let .
Then .
.
If , .
If , .
So, .
Substitute back : Given .
Comparing, and .
We need to find .
Since .
This matches option (3).

We know that the integral of is . To solve the given integral, we use substitution: Thus, when we perform the integration, we get: For the given options, option (3) simplifies to this result. Hence, the final solution is

Using substitution: Differentiating: Rewriting the integral: Applying the power rule: Another integral term: Rewriting: Thus, the correct answer is:

We are asked to evaluate: Step 1: Use integration by parts. Let Then and Step 2: Apply integration by parts formula. Step 3: Simplify second integral. We can write: Let’s denote Make substitution: Then and and This substitution gets messy — however, by known standard result: Step 4: Final integration result. Using the result:

Step 1: Interpret the integrand
The integral likely contains a typo, as is undefined. Assume , so the integral is , which matches the form of the options. Step 2: Integrate using substitution
Let , so , . Then , , and the integral becomes . Use integration by parts: , , , . Then . Continue: , and so on. Step 3: Compute and match
After three integrations by parts, the result is . Substitute back: , matching option (1).

Let .
Let .
Then .
Use partial fractions for .
Let .
Then .
.
If , .
If , .
So, .
Substitute back : Given .
Comparing, and .
We need to find .
Since .
This matches option (3).