`int(dx)/(2x^2+3x+5)=`

Options

A. `(2)/(sqrt(31))tan^(-1)((2x+3)/(sqrt(31)))+c`
B. `(4)/(sqrt(31))tan^(-1)((2x+3)/(sqrt(31)))+c`
C. `(2)/(sqrt(31))tan^(-1)((4x+3)/(sqrt(31)))+c`
D. `(4)/(sqrt(31))tan^(-1)((4x+3)/(sqrt(31)))+c`

Answer

To solve the integral \(\int \frac{dx}{2x^2 + 3x + 5}\), we will follow these steps:

### Step 1: Factor out the coefficient of \(x^2\)
We start by factoring out the coefficient of \(x^2\) from the denominator:
\[ \int \frac{dx}{2(x^2 + \frac{3}{2}x + \frac{5}{2})} \]
This can be rewritten as:
\[ \frac{1}{2} \int \frac{dx}{x^2 + \frac{3}{2}x + \frac{5}{2}} \]

**Hint:** Always factor out the leading coefficient when dealing with quadratic expressions in the denominator.

### Step 2: Complete the square
Next, we complete the square for the quadratic expression \(x^2 + \frac{3}{2}x + \frac{5}{2}\). To do this, we take half of the coefficient of \(x\) (which is \(\frac{3}{2}\)), square it, and add and subtract it inside the integral:
\[ \text{Half of } \frac{3}{2} = \frac{3}{4}, \quad \left(\frac{3}{4}\right)^2 = \frac{9}{16} \]
Thus, we can rewrite the quadratic as:
\[ x^2 + \frac{3}{2}x + \frac{9}{16} - \frac{9}{16} + \frac{5}{2} = \left(x + \frac{3}{4}\right)^2 - \frac{9}{16} + \frac{40}{16} = \left(x + \frac{3}{4}\right)^2 + \frac{31}{16} \]
Now, we can rewrite the integral:
\[ \frac{1}{2} \int \frac{dx}{\left(x + \frac{3}{4}\right)^2 + \frac{31}{16}} \]

**Hint:** Completing the square helps to transform the quadratic into a standard form that is easier to integrate.

### Step 3: Use the standard integral formula
We recognize that the integral now resembles the standard form:
\[ \int \frac{dx}{x^2 + a^2} = \frac{1}{a} \tan^{-1}\left(\frac{x}{a}\right) + C \]
In our case, \(a^2 = \frac{31}{16}\) which gives \(a = \frac{\sqrt{31}}{4}\). Therefore, we can rewrite our integral as:
\[ \frac{1}{2} \cdot \frac{4}{\sqrt{31}} \tan^{-1}\left(\frac{x + \frac{3}{4}}{\frac{\sqrt{31}}{4}}\right) + C \]

**Hint:** Recognizing standard integral forms allows for quicker integration.

### Step 4: Simplify the expression
Now, simplifying the expression:
\[ \frac{2}{\sqrt{31}} \tan^{-1}\left(\frac{4(x + \frac{3}{4})}{\sqrt{31}}\right) + C \]
This simplifies to:
\[ \frac{2}{\sqrt{31}} \tan^{-1}\left(\frac{4x + 3}{\sqrt{31}}\right) + C \]

**Hint:** Always simplify your final expression to its most concise form.

### Final Answer
Thus, the final result of the integral is:
\[ \int \frac{dx}{2x^2 + 3x + 5} = \frac{2}{\sqrt{31}} \tan^{-1}\left(\frac{4x + 3}{\sqrt{31}}\right) + C \]