How to compute the indefinite integral $\int \frac u{u+1}\,\mathrm du$?

Question

How do you compute $$ \int \frac{u}{u+1}\,\mathrm du$$

Answer 1

Hint:

Integrate by parts.

$$\int\dfrac{u}{u+1}du=u \int\dfrac{1}{u+1}du-\int \log(u+1)du$$

By the series of comments with Did,

$$\int \log u du= \log u\int1 du-\int u \cdot \dfrac{1}{u}du= (u\log u-u)$$

Answer 2

integrate by parts using : $f = u$ and $g' = 1/ (1+u)$. this solves the problem

Answer 3

$$ \int \frac{u}{u+1} du = \int \frac{u+1-1}{u+1} du = \int 1 \, du - \int \frac{1}{u+1} du. $$

Now it should be easy to solve.

Answer 4

Integration by parts needlessly complicates this problem.

A simpler approach uses partial fraction decomposition. Since $$\frac{u}{u+1} = 1-\frac{1}{u+1},$$ we have $$\int \frac{u}{u+1} du = \int 1du - \int \frac{1}{u+1}du = u - \log \vert u+1 \vert + c.$$