§(7x^2+5x+13/(x^2+2)(x+1)dx?

To solve this integral, we need to first decompose the rational function into partial fractions.

Given:

(7x^2 + 5x + 13) / ((x^2 + 2)(x + 1))

We can rewrite the numerator as:

7x^2 + 5x + 13 = Ax^2 + Bx + C

Multiplying both sides by (x^2 + 2)(x + 1), we get:

7x^2 + 5x + 13 = Ax^2(x + 1) + B(x^2 + 2) + C(x^2 + 2)(x + 1)

Expanding and equating coefficients, we get:

7x^2 + 5x + 13 = Ax^3 + Ax^2 + Bx^2 + 2B + Cx^3 + 2Cx^2 + 2C

Now, we can equate coefficients of like terms:

For x^3 term: A + C = 7
For x^2 term: A + B + 2C = 0
For x term: 2B = 5
For constant term: 2B + 2C = 13

Solving these equations, we get:

A = 1, B = 5/2, C = 6

Therefore, the partial fraction decomposition is:

(7x^2 + 5x + 13) / ((x^2 + 2)(x + 1)) = (1/(x + 1)) + (5/2(x^2 + 2)) + (6/(x^2 + 2))

Now, we can integrate each term separately:

∫(1/(x + 1)) dx = ln|x + 1| + C1
∫(5/2(x^2 + 2)) dx = (5/2)∫(x^2 + 2) dx = (5/2)(1/3)x^3 + 2x + C2 = (5/6)x^3 + 2x + C2
∫(6/(x^2 + 2)) dx = 6∫(1/(x^2 + 2)) dx = 6(1/√2)arctan(x/√2) + C3 = (3/√2)arctan(x/√2) + C3

Therefore, the final solution to the integral is:

ln|x + 1| + (5/6)x^3 + 2x + (3/√2)arctan(x/√2) + C

∫(7x^2 + 5x + 13)/(x^2 + 2x + 1) dx

Step 1: Factor the denominator

x^2 + 2x + 1 = (x + 1)^2

Step 2: Use partial fraction decomposition

7x^2 + 5x + 13 = A(x + 1)^2 + B(x + 1) + C

Equating coefficients:

- A = 7
- B = -5
- C = 13

Step 3: Rewrite the integrand

7x^2 + 5x + 13 = 7(x + 1)^2 - 5(x + 1) + 13

Step 4: Integrate

∫(7(x + 1)^2 - 5(x + 1) + 13)/(x + 1)^2 dx

For the term 7(x + 1)^2/(x + 1)^2:

= 7∫ dx
= 7x + C

For the term -5(x + 1)/(x + 1)^2:

Use the substitution u = x + 1, then du = dx

= -5∫ 1/u du
= -5ln|u| + C
= -5ln|x + 1| + C

For the term 13/(x + 1)^2:

= 13∫ 1/(x + 1)^2 dx
= -13/(x + 1) + C

Final answer:

∫(7x^2 + 5x + 13)/(x^2 + 2x + 1) dx = 7x - 5ln|x + 1| - 13/(x + 1) + C