How Can I Solve This Integration by Parts Problem?

[AStaunton]

problem is solve the following integral by parts:

$$\int\ln(2x+3)dx$$

I used substitution:

$$u=ln(2x+3)$$

$$\Rightarrow du=\frac{2}{2x+3}dx$$

and for dv:

$$dv=dx$$

$$\Rightarrow v=x$$

however, once I plug all these into my integration by parts formula, I get:

$$x\ln(2x+3)-\int\frac{2x}{2x+3}dx$$

and this new integral seems hard to solve...

Any advice appreciated on what went wrong.

Thanks

 

[Mark44]

You can approach this in either of two ways: polynomial long division or some fiddling with the numerator.

In the second approach, 2x/(2x + 3) = (2x + 3 - 3)/(2x + 3) = 1 - 3/(2x + 3), which is the same as what you would get using polynomial division.

 

[AStaunton]

thanks..is that called partial fractions?

 

[osnarf]

or you could do a substitution and let u = 2x + 3 for the bottom. Then, 2x = u - 3, and du = 2dx, so you're left with the integral of ((u - 3)/(2u))du

 

[osnarf]

Not quite, a partial fractions decomposition on it isn't really appropriate because the bottom doesn't break down anymore. It would be more appropriate in the case

(2x + 3)/(x * (2x + 1)), for which you would find A and B such that

(2x + 3)/(x * (2x + 1)) = (A/x) + (B/(2x + 1))

2x + 3 = A*( 2x + 1) + B*x

(2)*x + (3) = (2A + B)*x + (A)

Then,
A = 3
2A + B = 2

so
6 + B = 2
B = -4

Which means we can represent the original fraction as
(2/x) - (4/(2x + 1))

 

[SammyS]

thanks..is that called partial fractions?

No, as Mark44 said, it's "the same as what you would get using polynomial division".

Partial fractions is a way to decompose a fraction, when its denominator can be factored.