Can some one check my work on an integration problem?

Question

chrisplusian · Answer

$$\int\limits(\frac{ 1 }{ 7x-2 }-\frac{ 1 }{ 7x+2 })dx$$ thi is the original problem. I wan to show you the steps I followed, because when I check it with my ti nspire cas cx it gives a completely different answer.

chrisplusian · Answer

first$$\int\limits(\frac{ 1 }{7x-2 }dx -\int\limits \frac{ 1 }{ 7x+2 }dx$$

chrisplusian · Answer

Then for the first indefinite integral$$u=7x-2, du=7dx$$, then for the second indefinite integral$$u=7x+2, du=7dx$$

chrisplusian · Answer

then I have $$\frac{ 1 }{ 7 }\int\limits \frac{ 1 }{ u }du-\frac{ 1 }{ 7 }\int\limits \frac{ 1 }{ u }du$$

chrisplusian · Answer

finally my answer was$$\frac{ 1 }{ 7 }(\ln [7x-2]-\ln [7x+2])+c$$ those brackets should be absolute value bars but i couldn't figure out how to use them

hartnn · Answer

$\huge ln |A|-ln|B|=ln|\frac{A}{B}|$

chrisplusian · Answer

the cas calculator answer was $$\frac{ \ln \frac{ 7x-2 }{ 7x+2 } }{ 7 }$$

chrisplusian · Answer

so thats all I am missing is simplifying?

hartnn · Answer

$\huge ln |A|-ln|B|=ln|\frac{A}{B}|$