How do you integrate int x^3 ln x^2 dx  using integration by parts?

Oct 5, 2016

${x}^{4} / 8 \left(4 \ln x - 1\right) + C$

Explanation:

Let $I = \int {x}^{3} \ln {x}^{2} \mathrm{dx}$

Using the Subst. ${x}^{2} = t \text{, we have, } 2 x \mathrm{dx} = \mathrm{dt} , \mathmr{and} , x \mathrm{dx} = \frac{1}{2} \mathrm{dt} .$ Hence,

$I = \int \left({x}^{2} \ln {x}^{2}\right) x \mathrm{dx} = \frac{1}{2} \int t \ln t \mathrm{dt} .$

Now, to this end, we will use the Rule of I ntegration byParts arts :

$I B P : \int u v \mathrm{dt} = u \int v \mathrm{dt} - \int \left\{\frac{\mathrm{du}}{\mathrm{dt}} \int v \mathrm{dt}\right\} \mathrm{dt} .$

We take u=lnt rArr (du)/dt=1/t, &, v=trArr intvdt=t^2/2

$\therefore I = \frac{1}{2} \left[{t}^{2} / 2 \ln t - \int \left(\frac{1}{t} \cdot {t}^{2} / 2\right) \mathrm{dt}\right]$

$= {t}^{2} / 4 \ln t - \frac{1}{4} \int t \mathrm{dt} = {t}^{2} / 4 \ln t - {t}^{2} / 8$

Returning $t \text{ by } {x}^{2} ,$ we have,

$I = {x}^{4} / 4 \ln \left({x}^{2}\right) - \frac{1}{8} {x}^{4}$

Here, we use the Rule of Logarithm $: \ln {x}^{2} = 2 \ln x$ & get,

$I = {x}^{4} / 2 \ln x - {x}^{4} / 8 = {x}^{4} / 8 \left(4 \ln x - 1\right)$