How do you integrate 1/(x^2+3x+2)?

Any integration of the type 1/(ax^2 + bx + x) can always be done by the following steps.

ax^2 + bx + c

=a[x^2 + (b/a)x + c/a]

= a[(x + b/2a)^2 - b^2/(4a^2) + c/a]

Then use the standard formula of integration of 1 / (x^2 + a^2)

However, before using this sure method, examine if the denominator can be factored and resolved into partial fractions which is possible in your problem. This is simpler method. Use the general method described above only when factoring is not possible.

Thus, x^2 + 3x + 2 = (x +1)(x +2)

=> 1 / (x^2 + 3x +2) = 1 / (x +1) - 1 / (x + 2)

=> ∫ 1 / (x^2 + 3x + 2) dx

= ∫ 1 / (x +1) dx - ∫ 1 / (x + 2) dx

= ln l (x +1) / (x + 2) l + c

Now, let us see how to do your problem by the general method.

x^2 + 3x + 2

= [(x + 3/2)^2 -9/4 + 2]

= [(x + 3/2)^2 - (1/2)^2]

∫ 1 / (x^2 + 3x + 2) dx

= ∫ 1 / [(x + 3/2)^2 - (1/2)^2] dx

= 1 / [ 2 (1/2) ] * ln l (x + 3/2 - 1/2) / (x + 3/2 + 1/2) l + c

= ln l (x + 1) / (x + 2) l + c

Greetings,

Sorry why does partial fractions not work?

x^2 +3x + 2 = (x+1)(x+2)

1/(x^2+3x+2) = 1/(x+1) - 1/(x+2)

integral 1/(x^2+3x+2) dx = ln (x+1) - ln (x+2) + C

Regards

a. 1/(x^2+3x+2)=1/((x+2)(x+1) differention by parts

b. 1/(x^2+3x+2)=a/(x+1)+b/(x+2) multiply through by

X^2+3x+2 to clear fraction giving you

c. 1=a(x+2)+b(x+1)

d. 1=ax+2a+bx+b now group like factors together

e. x(a+b)+1(2a+b)=1

f. assign values from original functions

a+b=0

2a+b=1 multiply the first equation through by negative 1 and add together and you get B= -1 substitute and you get A=1

g. intergrate 1/(x+1)-1/(x+2)

h. y=ln|(x+1)|-ln|(x-2)|+c since subtraction implies division combine log functions

i. y=ln|(x+1)/(x+2)|+c

SARA!! you could either complete the square and use a u subs or PARTIAL FRACTIONS. Either way your answer will be ln of a quotient

Damn, I should know this. Sorry, I'm blanking as well.