Integration - I know where it went wrong, but not how.

[Glathberif]

Hello,

I've been trying to do the following integral (well, it's a only a step in a bigger integration by parts, but I know that the rest is right, but not that part):

INTEGRAL 3x/(9x²+12x+5)dx

Which I have factored into:

INTEGRAL (1/2-i)/(3x+2+i)dx + INTEGRAL (1/2+i)/(3x+2-i)dx

Now, according to Wolframalpha, those two previous integrals are equal, so the problem seems to be in this next step:
Since 1/2 and i are both constants, I take them out of the integral and I multiply by 3/3 (on will go in, one out):

1/(3/2-3i)* INTEGRAL 3/(3x+2+i)dx + 1/(3/2+3i)* INTEGRAL 3/(3x+2-i)dx

Which would give, since d(3x+2+i)/dx=3:

1/(3/2-3i)*ln(3x+2+i) + 1/(3/2+3i)*ln(3x+2-i)

Which is different from the answer given by Wolframalpha, and also doesn't derivate into the original function.

Some help would be appreciated, thanks!

 

[Deleted member 4993]

Hello,

I've been trying to do the following integral (well, it's a only a step in a bigger integration by parts, but I know that the rest is right, but not that part):

INTEGRAL 3x/(9x²+12x+5)dx

Which I have factored into:

INTEGRAL (1/2-i)/(3x+2+i)dx + INTEGRAL (1/2+i)/(3x+2-i)dx

Now, according to Wolframalpha, those two previous integrals are equal, so the problem seems to be in this next step:
Since 1/2 and i are both constants, I take them out of the integral and I multiply by 3/3 (on will go in, one out):

This does not make sense to me!!

the next step should be:

\(\displaystyle \displaystyle \frac{1}{3} * \left (\frac{1}{2} - i\right) * \int \frac{3 dx}{3*x + 2 - i} + \frac{1}{3} * \left (\frac{1}{2} + i\right) * \int \frac{3 dx}{3*x + 2 + i}\)

= \(\displaystyle \displaystyle \left (\frac{1 - 2*i}{6}\right) * \int \frac{3 dx}{3*x + 2 - i} + \left (\frac{1 + 2*i}{6}\right) * \int \frac{3 dx}{3*x + 2 + i}\)

= \(\displaystyle \displaystyle \left (\frac{1 - 2*i}{6}\right) * ln({3*x + 2 - i}) + \left (\frac{1 + 2*i}{6}\right) * ln({3*x + 2 + i}) + C\)

A simpler method would be to substitute (in the beginning)

u = 3*x + 2→ du = 3 dx

and end up with tan^-1(3*x+2) in the answer.



1/(3/2-3i)* INTEGRAL 3/(3x+2+i)dx + 1/(3/2+3i)* INTEGRAL 3/(3x+2-i)dx

Which would give, since d(3x+2+i)/dx=3:

1/(3/2-3i)*ln(3x+2+i) + 1/(3/2+3i)*ln(3x+2-i)

Which is different from the answer given by Wolframalpha, and also doesn't derivate into the original function.

Some help would be appreciated, thanks!

.

 

[Glathberif]

Okay, now that you've made the correction, I can finally see that I took the inverse of (1/2 +- i) when I took it out of the integral, which is so stupid I couldn't see it when I read it over to find the mistake. Sorry for making you lose your time for something like that.

 

[HallsofIvy]

A more standard method, since that denominator cannot be factored with real coefficients, would be to complete the square in the denominator: \(\displaystyle 9x^2+ 12x+ 5= 9(x^2+ \frac{4}{3}x+ \frac{5}{9})= 9(x^2+ \frac{4}{3}x+ \frac{4}{9}- \frac{4}{9}+ \frac{5}{9})\)
\(\displaystyle = 9((x+ \frac{2}{3})^2+ \frac{1}{9})\)

So \(\displaystyle \int\frac{3x}{9x^2+ 12x+ 5}dx=\)\(\displaystyle 9\int\frac{3x}{(x+ \frac{2}{3})^2+ \frac{1}{9}} dx\)

Let \(\displaystyle u= x+ \frac{2}{3}\) so that \(\displaystyle x= u- \frac{2}{3}\) and the integral becomes
\(\displaystyle 9\int \frac{u- \frac{2}{3}}{u^2+ \frac{1}{9}}dx= 9\int\frac{u}{u^2- \frac{2}{3}}dx- 6\int\frac{1}{u^2+ \frac{1}{9}}dx\).

The first can be integrated by letting \(\displaystyle v= u^2- \frac{2}{3}\) and the second by factoring \(\displaystyle \frac{1}{9}\) out of the denominator to get \(\displaystyle - 54\int\frac{1}{9u^2+ 1}dx\), using the substitution v= 3u, and then the integral is arctan(v)+ C.

I suspect that is the result Wolfram is giving.

 

[Glathberif]

Indeed, completing the square is better, but I didn't know about that method when I tried this integral.

So eventually, I got the correct answer (well, at least Wolframalpha says the difference between his answer and mine is 0) but with a term being a*ln(b) with (a,b) complex numbers instead of 2/3*arctan(2x+3), and it troubled me since I have no idea how to convert that into an arctan, so I derivated it and integrated it again and it worked.