# Can you evaluate int \ (xln(x+1))/(x+1) \ dx ?

May 16, 2017

I tried this:

#### Explanation:

Have a look:

May 16, 2017

$\int \setminus \frac{x \ln \left(x + 1\right)}{x + 1} \setminus \mathrm{dx} = \left(x + 1\right) \ln \left(x + 1\right) - x - \frac{1}{2} {\ln}^{2} \left(x + 1\right) + C$

#### Explanation:

Yes I can!

We want to find:

$I = \int \setminus \frac{x \ln \left(x + 1\right)}{x + 1} \setminus \mathrm{dx}$

An obvious logical substitution would be:

Let $w = x + 1 = > \frac{\mathrm{dw}}{\mathrm{dx}} = 1$, and $x = w = 1$

Substituting into the integral gives us:

$I = \int \setminus \frac{\left(w - 1\right) \ln w}{w} \setminus \mathrm{dw}$
$\setminus \setminus = \int \setminus \frac{w \ln w - \ln w}{w} \setminus \mathrm{dw}$
$\setminus \setminus = \int \setminus \ln w - \ln \frac{w}{w} \setminus \mathrm{dw}$
$\setminus \setminus = \int \setminus \ln w \setminus \mathrm{dw} - \int \setminus \ln \frac{w}{w} \setminus \mathrm{dw}$

The first integral is a well known result and one that probably should be memorised. It can be derived using Integration by parts:

Let $\left\{\begin{matrix}u & = \ln w & \implies & \frac{\mathrm{du}}{\mathrm{dw}} = \frac{1}{w} \\ \frac{\mathrm{dv}}{\mathrm{dw}} & = 1 & \implies & v = w\end{matrix}\right.$

Then plugging into the IBP formula:

$\int \setminus \left(u\right) \left(\frac{\mathrm{dv}}{\mathrm{dx}}\right) \setminus \mathrm{dx} = \left(u\right) \left(v\right) - \int \setminus \left(v\right) \left(\frac{\mathrm{du}}{\mathrm{dx}}\right) \setminus \mathrm{dx}$

gives us:

$\int \setminus \left(\ln w\right) \left(1\right) \setminus \mathrm{dw} = \left(\ln w\right) \left(w\right) - \int \setminus \left(w\right) \left(\frac{1}{1}\right) \setminus \mathrm{dw}$
$\therefore \int \setminus \ln w \setminus \mathrm{dw} = w \ln w - \int \setminus \mathrm{dw}$
$\therefore \text{ } = w \ln w - w$

And for the second integral we can perform another substitution:

Let $z = \ln w \implies \frac{\mathrm{dz}}{\mathrm{dw}} = \frac{1}{w}$

Substituting into the second integral gives us:

$\int \setminus \ln \frac{w}{w} \setminus \mathrm{dw} = \int \setminus z \setminus \mathrm{dz}$
$\text{ } = \frac{1}{2} {z}^{2}$

And restoring the substitution for $w$ we get

$\int \setminus \ln \frac{w}{w} \setminus \mathrm{dw} = \frac{1}{2} {\left(\ln w\right)}^{2}$

Combining these two results then gives us:

$I = \left\{w \ln w - w\right\} - \left\{\frac{1}{2} {\left(\ln w\right)}^{2}\right\} + {C}_{1}$

And restoring the substitution for $w$ gives us:

$I = \left(x + 1\right) \ln \left(x + 1\right) - \left(x + 1\right) - \frac{1}{2} {\ln}^{2} \left(x + 1\right) + {C}_{1}$
$\setminus \setminus = \left(x + 1\right) \ln \left(x + 1\right) - x - 1 - \frac{1}{2} {\ln}^{2} \left(x + 1\right) + {C}_{1}$
$\setminus \setminus = \left(x + 1\right) \ln \left(x + 1\right) - x - \frac{1}{2} {\ln}^{2} \left(x + 1\right) + C$