Question

Evaluate \( \displaystyle \int \frac{x+2}{2x^2 + 6x + 5} \, dx \) using the substitution or partial fraction method.

Hint:

If numerator is close to derivative of denominator, split it into derivative form + remainder to simplify integration.

Answer 1

To Evaluate:
∫ (x + 2) / (2x² + 6x + 5) dx

Step 1: Observe the Denominator
Let D = 2x² + 6x + 5

Differentiate D:

dD/dx = 4x + 6

Now rewrite the numerator (x + 2) in terms of (4x + 6).

Express x + 2 as:

x + 2 = A(4x + 6) + B

Comparing coefficients:

x + 2 = 4A x + 6A + B

Equating coefficients:

4A = 1 ⇒ A = 1/4

6A + B = 2 
6(1/4) + B = 2 
3/2 + B = 2 
B = 1/2

So,

x + 2 = (1/4)(4x + 6) + 1/2

Step 2: Split the Integral

∫ (x+2)/(2x²+6x+5) dx =

(1/4) ∫ (4x+6)/(2x²+6x+5) dx + (1/2) ∫ 1/(2x²+6x+5) dx

Step 3: First Integral

Let u = 2x² + 6x + 5

du = (4x + 6) dx

So,

(1/4) ∫ du/u = (1/4) ln|2x² + 6x + 5|

Step 4: Second Integral

Complete the square:

2x² + 6x + 5 = 2(x² + 3x) + 5

= 2[(x + 3/2)² − 9/4] + 5

= 2(x + 3/2)² − 9/2 + 5

= 2(x + 3/2)² + 1/2

= 2[(x + 3/2)² + 1/4]

Therefore,

(1/2) ∫ dx / [2((x + 3/2)² + (1/2)²)]

= (1/4) ∫ dx / [(x + 3/2)² + (1/2)²]

Using standard formula:

∫ dx/(x² + a²) = (1/a) tan⁻¹(x/a)

So,

(1/4) × (1/(1/2)) tan^{−1>((x + 3/2)/(1/2))}

= (1/2) tan^{−1>(2x + 3)}

Final Answer:
∫ (x+2)/(2x²+6x+5) dx =

(1/4) ln|2x² + 6x + 5| + (1/2) tan^{−1>(2x + 3) + C}