# I really need help with this plz i) Find d/dx(x^3lnx) ii) Hence find intx^2lnxdx ?

## i,ve done the first part but i cant do the second very badly stuck

Mar 22, 2018

I get $\frac{{x}^{3} \left(3 \ln x - 1\right)}{9} + C$.

#### Explanation:

Given: $\int {x}^{2} \ln x \setminus \mathrm{dx}$

Apply integration by parts, $\int u \setminus \mathrm{dv} = u v - \int v \setminus \mathrm{du}$.

Here, I'll let $u = \ln x , \mathrm{dv} = {x}^{2}$

$\therefore \mathrm{du} = \frac{1}{x} \setminus \mathrm{dx} , v = \frac{1}{3} {x}^{3}$ (no need $C$ yet)

So, going back to our original integral, we get

$\int \left({x}^{2} \ln x\right) \setminus \mathrm{dx} = \ln x \cdot \frac{1}{3} {x}^{3} - \int \left(\frac{1}{3} {x}^{3} \cdot \frac{1}{x}\right) \setminus \mathrm{dx}$

$= \frac{{x}^{3} \ln x}{3} - \int \left({x}^{2} / 3\right) \setminus \mathrm{dx}$

$= \frac{{x}^{3} \ln x}{3} - \frac{1}{3} \int {x}^{2} \setminus \mathrm{dx}$

$= \frac{{x}^{3} \ln x}{3} - \frac{1}{3} \cdot {x}^{3} / 3 + C$

$= \frac{{x}^{3} \ln x}{3} - {x}^{3} / 9 + C$

$= \frac{3 {x}^{3} \ln x - {x}^{3}}{9} + C$

$= \frac{{x}^{3} \left(3 \ln x - 1\right)}{9} + C$

Mar 22, 2018

$\frac{d}{\mathrm{dx}} \left({x}^{3} \ln x\right) = {x}^{2} + 3 {x}^{2} \ln x$

$\int \setminus {x}^{2} \ln x \setminus \mathrm{dx} = \frac{1}{3} {x}^{3} \ln x - {x}^{3} / 9 + C$

#### Explanation:

Part (i)

We seek:

$\frac{d}{\mathrm{dx}} \left({x}^{3} \ln x\right)$

We can use the product rule, to get:

 d/dx (x^3lnx) = (x^3)(d/dxlnx))+(d/dxx^3)(lnx)
$\text{ } = \left({x}^{3}\right) \left(\frac{1}{x}\right) + \left(3 {x}^{2}\right) \left(\ln x\right)$
$\text{ } = {x}^{2} + 3 {x}^{2} \ln x$

Part (ii)

"Hence", indicates we should probably use the earlier result:

$\frac{d}{\mathrm{dx}} \left({x}^{3} \ln x\right) = {x}^{2} + 3 {x}^{2} \ln x$

Integrate (term by term) wrt $x$ (omitting the constant of integration), we get

${x}^{3} \ln x = \int \setminus {x}^{2} \setminus \mathrm{dx} + \int \setminus 3 {x}^{2} \ln x \setminus \mathrm{dx}$

The first integral is trivial (power rule), the second integral is the result we seek, thus:

${x}^{3} \ln x = {x}^{3} / 3 + 3 \setminus \int \setminus {x}^{2} \ln x \setminus \mathrm{dx}$

And we can rearrange:

$3 \setminus \int \setminus {x}^{2} \ln x \setminus \mathrm{dx} = {x}^{3} \ln x - {x}^{3} / 3$

And incorporating the constant of integration:

$\int \setminus {x}^{2} \ln x \setminus \mathrm{dx} = \frac{1}{3} {x}^{3} \ln x - {x}^{3} / 9 + C$