How do you integrate #xln(1+x) dx#?

Question

Cameron Alexander · Accepted Answer

The solution is shown here (sometimes Wolfram Alpha has trouble  computing integrals with #lnx# properly, so I gave the backwards check). I would separate the integral and begin with a u-substitution. Let:
#u = x+1#
#du = dx#
#x = u - 1# #=> int (u-1)lnudu# #= int u lnudu - int lnudu# With these two integrals in mind, we can do Integration by Parts (assuming you already know the integral of #lnx#). Ignoring the #int lnu#, let: #s = lnu#
#ds = 1/udu#
#dt = udu#
#t = u^2/2# Thus: #st - int tds# #= [(u^2lnu)/2 - int u^2/2*1/udu] - int lnudu# #= (u^2lnu)/2 - 1/2int udu - int lnudu# #= (u^2lnu)/2 - u^2/4 - (u lnu - u)# #= (u^2lnu)/2 - u^2/4 - u lnu + u# #= (u^2lnu)/2 - u lnu - u^2/4 + u# #= ((x+1)^2ln(x+1))/2 - (x+1) ln(x+1) - (x+1)^2/4 + x + 1 + C# And #1# gets embedded into #C#: #= color(blue)(((x+1)^2ln(x+1))/2 - (x+1) ln(x+1) - (x+1)^2/4 + x + C)#

Christopher Allers · Answer

undefined To integrate $x \ln(1+x) \, dx$, you can use integration by parts. Let $u = \ln(1+x)$ and $dv = x \, dx$. Then, $du = \frac{1}{1+x} \, dx$ and $v = \frac{1}{2}x^2$. Applying the integration by parts formula $\int u \, dv = uv - \int v \, du$, we have:

$$
\begin{align*}
\int x \ln(1+x) \, dx &= \frac{1}{2}x^2 \ln(1+x) - \int \frac{1}{2}x^2 \cdot \frac{1}{1+x} \, dx \
&= \frac{1}{2}x^2 \ln(1+x) - \frac{1}{2} \int \frac{x^2}{1+x} \, dx
\end{align*}
$$

Now, to integrate $\int \frac{x^2}{1+x} \, dx$, use polynomial division or rewrite it as $\int \left(x - \frac{x}{1+x}\right) \, dx$. Then, integrate term by term:

$$
\int \frac{x^2}{1+x} \, dx = \int \left(x - \frac{x}{1+x}\right) \, dx
$$

$$
= \frac{1}{2}x^2 - \int \frac{x}{1+x} \, dx
$$

For the integral $\int \frac{x}{1+x} \, dx$, perform a simple substitution or use $u$-substitution with $u = 1+x$. This results in:

$$
\int \frac{x}{1+x} \, dx = \frac{1}{2}x^2 - \ln|1+x| + C
$$

Substitute this back into the original equation:

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

Simplify this expression to obtain the final answer.