Question

Use partial fraction to evaluate the following integral: 29 6x9 10x3 3dx x2 6x 10

A) 29ln∣x²+6x+10∣+C
B) 29ln∣x+3∣+6ln∣x+2∣+C
C) 29ln∣x+3∣+C
D) 29ln∣x+2∣+6ln∣x+5∣+C
E) 29ln∣x+2∣+C

Answer 1

To evaluate the given integral using partial fractions, we factor the denominator into irreducible quadratic factors and decompose the fraction. After equating coefficients, we can evaluate each integral separately and obtain the final result of 6ln|x + 3| + 23ln|x + 2| + C.

Step-by-step explanation:

To evaluate the given integral, we can use partial fractions. First, we factor the denominator into irreducible quadratic factors, which are x^2 + 6x + 10 = (x + 3)(x + 2). Then we decompose the fraction into its partial fraction form:

(29x^3 + 6x^2 + 3) / ((x + 3)(x + 2)) = A / (x + 3) + B / (x + 2)

Multiplying through by the denominator and simplifying, we have:

29x^3 + 6x^2 + 3 = A(x + 2) + B(x + 3)

Expanding and equating coefficients, we find A = 6 and B = 23.

Therefore, the integral becomes:

∫(29x^3 + 6x^2 + 3) / ((x + 3)(x + 2)) dx = ∫(6 / (x + 3)) dx + ∫(23 / (x + 2)) dx

Using the integral rules, we can evaluate each integral separately:

∫(6 / (x + 3)) dx = 6ln|x + 3|

∫(23 / (x + 2)) dx = 23ln|x + 2|

Therefore, the final result is: 6ln|x + 3| + 23ln|x + 2| + C.