2 different integrals

Question

∫x/(X² +3)  = .5 ln(x²+3)    

 

u=x²+3

du = 2x dx

 

1/2 ∫ 1/(x²+3)  =  1/2 *ln(x2+3)

 

here ∫ 1/(x2+3) gives is ln(

 

https://www.symbolab.com/solver/definite-integral-calculator/%5Cint%5Cfrac%7Bx%7D%7B%5Cleft%28x%5E%7B2%7D%2B3%5Cright%29%7D/?origin=button

 

But when computing this integral alone   ∫ 1/(x2+3)

you get something to do with arc tan

 

https://www.symbolab.com/solver/definite-integral-calculator/%5Cint%5Cfrac%7B1%7D%7B%5Cleft%28x%5E%7B2%7D%2B3%5Cright%29%7D/?origin=button

 

why?

 

 

Answer 1

Because this line:

1/2 ∫ 1/(x²+3) = 1/2 *ln(x2+3)

should look like this after the u substitution:

1/2 ∫ (1/u)du = 1/2 ln |u|, now substitute back in the x² + 3: 1/2 ln(x²+3) + C

Note the absolute value is dropped in the final answer because x² + 3 has a minimum value of 3 > 0.

Note that the antiderivative of 1/(x² + 1) is indeed arctan(x).