Question

integrate:(x+3)/sqrt(x^2+4x+5)

Answer 1

I would try to factor the quadratic or something.

Answer 2

u have to firs express numerator as A(diif of denom )+B where A and B are constatnts to be dtermined
thus A(2x+4) +B =x+3

Answer 3

After you factor the denominator you will get an expression that lends itself (not too nicely) to a few layers of substitution. I am getting a rather complicated trigonometric answer when I do it.

Answer 4

equating coefficient on both sides we have 2A=1 or A=1/2
and 4A+B=3 or B=1
thus (x+3)= 1/2 (2x+4) +1
hence reqd integ breaks down to two as 1/2 (2x+4)/(x^2+4x+5) +1 /((x^2+4x+5)
let I =integral of 1/2 (2x+4)/(x^2+4x+5) and J = integral of 1 /((x^2+4x+5)


now one has to eavaluate first I and J

Answer 5

Actually looks like you have to complete the square of the denominator.

Answer 6

Right, after completing the square on the bottom you can obtain an equivalent form where u = x + 2:

integrate (u + 1) / sqrt(u^2 + 1) du

Still complicated from here though.

Answer 7

\[(a+b)^2 = a^2 +2ab + b^2 \\
a = x \\
b = ?
\]

Answer 8

for I substitute (x^2+4x+5)=z
or (2x+4)dx=dz
hence I =1/2 (inegral of dz/z) =1/2 lnz +c1
or I=1/2 ln (x^2+4x+5)+c1

Answer 9

Now for J one has to complete sqaures
thus (x^2+4x+5)=(x^2+4x+4+1)=(x+2)^2 +1

Answer 10

@drake_96 You there?

Answer 11

J= integral of 1/((x+2)^2 +1) = tan^-1(x+2) +c2

Answer 12

thus reqd answer is
1/2 ln (x^2+4x+5) + tan^-1(x+2) +c [ where c=c1+c2 all integrating constant]

Answer 13

I think the derivative of the function you provided yields (x + 3) / (x^2 + 4x + 5) instead of (x + 3) / sqrt(x^2 + 4x + 5).

I got sqrt(x^2 + 4x + 5) + ln (x + 2 + sqrt( (x + 2)^2 + 1)) for the final answer. Can you double check this?

Answer 14

@LogicalApple :you are right.