Question

9.
Evaluate
x. logx. dx
​

Answer 1

Evaluate the following integral:

\[ \int x \log x \, dx. \]

We use integration by parts, defining

\begin{align*} u &= \log x & du &= \frac{1}{x} \, dx \\ dv &= x \, dx & v &= \frac{1}{2} x^2. \end{align*}

Then we have,

Answer 2

x log x dx = (x^2/2) log x -x^2/4 + c

Proof : Using integration by parts,

∫udv = uv - ∫vdu

In ∫ x log x dx,

take, u=logx => du= (1/x) . dx

∫dv=∫x dx => v=x^2/2

Now substituting,

∫x log x = logx (x^2/2) - ∫x ^/2 . (1/x) dx = logx (x^2/2) - (1/2)∫dx= (x^2/2) log x -x^2/4 + c

where c is constant.

Cheers,