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What is the integral of #xln(x)#?

Answer 1

Paisley Johnson

You have to use the Integration by Parts formula: #intudv=uv-intvdu#

let u= lnx du = #1/x#

dv = x v = #x^2/2#

Enter this into the IBP formula to obtain.

=#lnx*x^2/2 - intx^2/2*1/x# Solve the integral on the right side and you'll get #-x^2/4#

Final answer would be: #lnx*x^2/2-x^2/4#

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Answer 2

Luke Alvarez

The integral of ( x \ln(x) ) with respect to ( x ) can be found using integration by parts. Applying integration by parts, we let ( u = \ln(x) ) and ( dv = x , dx ). Then, ( du = \frac{1}{x} , dx ) and ( v = \frac{x^2}{2} ).

[ \begin{align*} \int x \ln(x) , dx &= x \cdot \frac{x^2}{2} - \int \frac{x^2}{2} \cdot \frac{1}{x} , dx \ &= \frac{x^3}{2} - \frac{1}{2} \int x , dx \ &= \frac{x^3}{2} - \frac{1}{2} \cdot \frac{x^2}{2} + C \ &= \frac{x^3}{2} - \frac{x^2}{4} + C \end{align*} ]

Therefore, the integral of ( x \ln(x) ) with respect to ( x ) is ( \frac{x^3}{2} - \frac{x^2}{4} + C ), where ( C ) is the constant of integration.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.