∫ (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.
∫ (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`
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.
To integrate the function ∫ (2x+1)/(x+1) dx, you are correct that you can use partial fraction decomposition. Here are the steps to solve it:
1. Start by writing the integrand as (2x+1)/(x+1).
2. Perform long division or synthetic division to divide (2x+1) by (x+1). This will give you a quotient, 2, and a remainder, -1.
(2x+1)/(x+1) = 2 - 1/(x+1)
3. Now, you can rewrite the integral as ∫ 2 dx - ∫ 1/(x+1) dx.
4. The first integral, ∫ 2 dx, is straightforward to solve. Integrating a constant, like 2, gives you 2x.
5. For the second integral, ∫ 1/(x+1) dx, you can use a substitution. Let u = x + 1, so du = dx. Rewriting the integral, you get ∫ 1/u du.
6. The integral of 1/u with respect to u is ln|u|. So, the second integral becomes ∫ 1/(x+1) dx = ln|x+1|.
7. Combining the results from steps 4 and 6, the original integral simplifies to 2x - ln|x+1| + C, where C is the constant of integration.
Therefore, the solution to the integral ∫ (2x+1)/(x+1) dx is 2x - ln|x+1| + C.