Integration problem

[Stef95]

I need help with this integral. I don't know how to start.

 

[MarkFL]

Hello, and welcome to FMH!

Try writing the integral as:

[MATH]\frac{1}{2}\int 1+\frac{1}{2ax+1}\,dx[/MATH]
Can you proceed?

 

[Stef95]

Thanks.
I will try to solve now

 

[MarkFL]

What I would do next is let:

[MATH]u=2ax+1\implies du=2a\,dx[/MATH]
And so now we have:

[MATH]\frac{1}{4a}\int 1+\frac{1}{u}\,dx=\frac{1}{4a}(u+\ln(u))+C[/MATH]
Now back substitute for \(u\):

[MATH]\int\frac{ax+1}{2ax+1}\,dx=\frac{1}{4a}(2ax+\ln(2ax+1))+C[/MATH]
Note: the constant that resulted in the back-substitution was "absorbed" into the constant of integration.

 

[lookagain]

And so now we have:

The natural logarithms in the antiderivatives get used with the absolute
value bars.

[MATH] ... \ = \ \frac{1}{4a}(u+\ln|u|) \ + \ C[/MATH]
Now back substitute for \(u\):

[MATH]\int\frac{ax+1}{2ax+1}\,dx \ = \ \frac{1}{4a}(2ax+\ln|2ax+1|) \ + \ C[/MATH]