The integration of x/(1+x), that is, the integral of x divided by 1+x is equal to x – ln|1+x|+C. In this post, we will learn how to integrate x/(1+x).

To find the integral $\int \dfrac{x}{1+x} dx$, we will use the following power rule of the integration formula:

∫dx/x =ln|x|+C **…(I)**

## Integration of x/(1+x)

**Question:** Find the integral

$\int \dfrac{x}{1+x} dx$.

**Solution:**

To find the integral of x/(1+x), first, we will add and subtract 1 from the numerator. By doing so we get that

$\dfrac{x}{1+x}$ = $\dfrac{(1+x)-1}{1+x}$

Rewriting, the above will be

= $\dfrac{1+x}{1+x} – \dfrac{1}{1+x}$

= $1 – \dfrac{1}{1+x}$

So we deduce that

$\dfrac{x}{1+x}$ = $1 – \dfrac{1}{1+x}$

Integrating both sides with respect to x, we get that

$\int \dfrac{x}{1+x} dx$ = $\int \left[1 – \dfrac{1}{1+x} \right] dx$

= ∫ 1 dx – ∫ $\dfrac{dx}{1+x}$

= ∫ 1 dx – ∫ dz/z where z=1+x.

= x – ln|1+x|+C. Here we have used the above formula **(I)** and have put the value of z. Here C is an integration constant.

**So the integration of x by 1+x is equal to x – ln|1+x|+C.**

**Have You Read These?**

Integration of Square Root of x

## FAQs

### Q1: What is the integration of x/(1+x)?

**Answer:** The integration of x/(1+x) is x – ln|1+x|+C. Here ln denotes the natural logarithm and C denotes an integral constant.