# Not seeing how to integrate y/(y+1)

1. Jan 27, 2006

### opticaltempest

Hello,

I cannot figure out how to evaluate the following integral,

$$\int {\frac{y}{{y + 1}}dy}$$

If it was $$y^2+1$$ then I see how a u-substitution of $$u=y^2+1$$ would work.

Thanks

2. Jan 27, 2006

### StatusX

Usually with fractions like this, you'll find you can express what you have as the sum of two or more fractions with the same denominator which are easier to integrate. In this case, try the numerators y+1 and -1.

3. Jan 27, 2006

### Benny

As in, write y = y + 1 - 1.

4. Jan 27, 2006

### opticaltempest

Ok, that is what I was having trouble seeing. Thanks

$$\frac{y}{{y + 1}} = \frac{{y + 1}}{{y + 1}} - \frac{1}{{y + 1}} = 1 - \frac{1}{{y + 1}}$$

$$\int {1{\rm }dy - } \int {\frac{1}{{y + 1}}{\rm }} dy = y - \ln \left| {y + 1} \right| + C$$

5. Jan 27, 2006

### 0rthodontist

You could also substitute u = y + 1, du = dy, and get the integral of (u - 1) / u with respect to u.