Question

integration of x log x​

Answer 1

∫x log x dx = (x²/2) log x - x²/4 + c

Step-by-step explanation:

• We know that

∫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 (x²/2) - ∫x/2 . (1/x) dx

= logx (x²/2) - (1/2)∫dx

= (x²/2) log x - x²/4 + c

where c is constant.

Answer 2

Answer:

$\sf I= \dfrac{x^2\times log\:x}{2} -\dfrac{x^2}{4} +C$

Step-by-step explanation:

Given:

x log x

To Find:

$\displaystyle \sf \int\limits {x\: log\:x} \, dx$

Solution:

$\displaystyle \sf \int\limits { x\:log\:x} \, dx$

Integrating by parts and using the ILATE rule,

We know that,

$\displaystyle \sf \int\limits {u.v} \, dx =u\int\limits{v} \, dx -\int\limits {\bigg(\dfrac{d}{dx}(u)\int\limits {v} \, dx } \bigg)\, dx$

where u is the first and v is the second function.

Here first function u = log x and second function v = x.

Hence,

$\displaystyle \sf \int\limits { x\:log\:x} \, dx=log\:x\: \int\limits {x} \, dx -\int\limits{\bigg(\dfrac{d}{dx}(log\:x)\: \int\limits {x} \, dx } \bigg)\, dx$

$\implies \displaystyle \sf log\:x\times \dfrac{x^2}{2} -\int\limits {\bigg(\dfrac{1}{x}\times \dfrac{x^2}{2} } \bigg)\, dx$

$\implies \displaystyle \sf log\:x\times \dfrac{x^2}{2} -\int\limits {\bigg( \dfrac{x}{2} } \bigg)\, dx$

$\implies \displaystyle \sf log\:x\times \dfrac{x^2}{2} -\dfrac{1}{2} \int\limits { x } \, dx$

$\implies \displaystyle \sf log\:x\times \dfrac{x^2}{2} -\dfrac{1}{2} \times \dfrac{x^2}{2} +C$

$\implies \sf \dfrac{x^2\times log\:x}{2} -\dfrac{x^2}{4} +C$

This is the integral of the given function.