What is the antiderivative of $\frac{ln (x^{3})}{x}$ $ln(x^3)/x$ ?

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What is the antiderivative of $\frac{ln (x^{3})}{x}$ $ln(x^3)/x$ ?

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Answer 1 · 2016-12-03T21:53:11

$\frac{3}{2} {(ln (x))}^{2} + C$ $3/2(ln(x))^2+C$

Explanation:

This is the same as asking:

$\int \frac{ln (x^{3})}{x} d x$ $intln(x^3)/xdx$

Using the logarithm rule $log (a^{b}) = b log (a)$ $log(a^b)=blog(a)$ we can move the constant out. Then, a good substitution would be $u = ln (x)$ $u=ln(x)$ , which implies that $d u = \frac{1}{x} d x$ $du=1/xdx$ (see this through taking the derivative of $ln (x)$ $ln(x)$ ).

$= 3 \int \frac{ln (x)}{x} d x = 3 \int ln (x) (\frac{1}{x} d x) = 3 \int u d u = 3 \frac{u^{2}}{2} = \frac{3}{2} {(ln (x))}^{2} + C$ $=3intln(x)/xdx=3intln(x)(1/xdx)=3intudu=3u^2/2=3/2(ln(x))^2+C$

We see that integration by parts isn't even necessary!

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What is the antiderivative of $\frac{ln (x^{3})}{x}$ $ln(x^3)/x$ ?

1 Answer

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