How do I find the integral $\int (x \cdot ln (x)) d x$ $int(x*ln(x))dx$ ?

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How do I find the integral $\int (x \cdot ln (x)) d x$ $int(x*ln(x))dx$ ?

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Answer 1 · 2014-08-18T03:03:54

We will use integration by parts.

Remember the IBP's formula, which is

$\int u d v = u v - \int v d u$ $int u dv = uv - int v du$

Let $u = ln x$ $u = ln x$ , and $d v = x d x$ $dv = x dx$ . We have chosen these values because we know that the derivative of $ln x$ $ln x$ is equal to $\frac{1}{x}$ $1/x$ , meaning that instead of integrating something complex (a natural logarithm) we now will end up integrating something pretty easy. (a polynomial)

Thus, $d u = \frac{1}{x} d x$ $du = 1/x dx$ , and $v = \frac{x^{2}}{2}$ $v = x^2 / 2$ .

Plugging into the IBP's formula gives us:

$\int x ln x d x = \frac{x^{2} ln x}{2} - \int \frac{x^{2}}{2 x} d x$ $int x ln x dx = (x^2 ln x)/2 - int x^2 / (2x) dx$

An $x$ $x$ will cancel off from the new integrand:

$\int x ln x d x = \frac{x^{2} ln x}{2} - \int \frac{x}{2} d x$ $int x ln x dx = (x^2 ln x)/2 - int x / 2 dx$

The solution is now easily found using the power rule. Don't forget the constant of integration:

$\int x ln x d x = \frac{x^{2} ln x}{2} - \frac{x^{2}}{4} + C$ $int x ln x dx = (x^2 ln x)/2 - x^2 / 4 + C$

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How do I find the integral $\int (x \cdot ln (x)) d x$ $int(x*ln(x))dx$ ?

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