Question

∫ (2x+1)/(x+1) dx

Please show steps if you work it. I think it's a partial fraction and worked with decomposition. Not sure though.

Answer 1

∫ (2x+1)/(x+1) dx
We can separate this term into two terms by taking each term in the numerator:
∫ (2x)/(x+1) + 1/(x+1) dx

For the term on the right,
∫ 1/(x+1) dx = ln |x+1|

For the term on the left,
Let u = x+1, thus x = u-1, and dx = du
Replacing,
2 ∫ (x)/(x+1) dx
2 ∫ (u-1)/(u) du
2 ∫ (u/u - 1/u) du
2 ∫ (1 - 1/u) du
= 2 ( u - ln|u| )
= 2(x+1) - 2*ln|x+1|

Combining, the integral is
= 2(x+1) - 2*ln|x+1| + ln |x+1| + C
= 2x + 2 - ln|x+1| + C

hope this helps~ `u`

Answer 2

you can use partial fractions.

(2x+1)/(x+1) = 2 - 1/x+1
so, the integral is

2x - ln(x+1) + C

Note that my C is different from Jai's, because it includes the extra 2.