# How do you integrate int x^3 ln x^4 dx  using integration by parts?

Mar 1, 2018

$\int {x}^{3} \ln \left({x}^{4}\right) \mathrm{dx} = {x}^{4} \ln \left(x\right) - \frac{1}{4} {x}^{4} + C$

#### Explanation:

First, let's rewrite $\ln \left({x}^{4}\right)$ as $4 \ln \left(x\right) ,$ as $\ln \left({x}^{a}\right) = a \ln \left(x\right)$

$\int {x}^{3} \ln \left({x}^{4}\right) \mathrm{dx} = \int 4 {x}^{3} \ln \left(x\right) \mathrm{dx} = 4 \int {x}^{3} \ln \left(x\right) \mathrm{dx}$ (We can factor out $4 ,$ it's just a constant.)

Make the following choices for $u$ and $\mathrm{dv} :$

$u = \ln \left(x\right)$

$\mathrm{dv} = {x}^{3} \mathrm{dx}$

Thus, $\mathrm{du}$ and $v$ become:

$\mathrm{du} = \frac{1}{x} \mathrm{dx}$

$\mathrm{dv} = \int {x}^{3} \mathrm{dx} = \frac{1}{4} {x}^{4}$ (We're not going to include the constant just yet. It can wait until the end.)

Now it becomes apparent why we made the choices we did for $u$ and $v .$ If we chose $\mathrm{dv} = \ln \left(x\right) ,$ we would have to integrate $\ln \left(x\right)$ for $v ,$ and would run into our original problem.

Plug in relevant values:

$u v - \int v \mathrm{du} = \frac{1}{4} {x}^{4} \ln \left(x\right) - \int \frac{1}{4} {x}^{4} / x \mathrm{dx}$

We can simplify the integral here:

$\frac{1}{4} {x}^{4} \ln \left(x\right) - \int \frac{1}{4} {x}^{4} / x \mathrm{dx} = \frac{1}{4} {x}^{4} \ln \left(x\right) - \frac{1}{4} \int {x}^{\cancel{4} 3} / \cancel{x}$

Thus, we have

$\frac{1}{4} {x}^{4} \ln \left(x\right) - \frac{1}{4} \int {x}^{3} \mathrm{dx} = \frac{1}{4} {x}^{4} \ln \left(x\right) - \left(\frac{1}{4}\right) \left(\frac{1}{4}\right) {x}^{4} + C = \frac{1}{4} {x}^{4} \ln \left(x\right) - \frac{1}{16} {x}^{4} + C$

Let's not forget the $4$ we originally factored out...

$4 \left(\frac{1}{4} {x}^{4} \ln \left(x\right) - \frac{1}{16} {x}^{4} + C\right) = {x}^{4} \ln \left(x\right) - \frac{1}{4} {x}^{4} + C$

$C$ remains unchanged; a constant multiplied by a constant remains a constant.

So,

$\int {x}^{3} \ln \left({x}^{4}\right) \mathrm{dx} = {x}^{4} \ln \left(x\right) - \frac{1}{4} {x}^{4} + C$

Mar 1, 2018

$\int \setminus {x}^{3} \ln \left({x}^{4}\right) \setminus \text{d} x = 4 {x}^{4} \ln \left(x\right) - {x}^{4} + C$

#### Explanation:

Here's another approach:

Consider substituting first $w = {x}^{4}$ and $\left(\text{d"w)/("d} x\right) = 4 {x}^{3}$ into the integral $\int \setminus {x}^{3} \ln \left({x}^{4}\right) \setminus \text{d} x$.

This gives $\frac{1}{4} \int \setminus \ln \left(w\right) \setminus \text{d} w$.

Now use integration by parts $\int \setminus u \setminus \text{d"v=uv-int\ v\ "d} u$. Use $u = \ln \left(w\right)$ and $\text{d"v="d} w$. Then find that "d"u=("d"w)/w and $v = w$.

Now, we have 1/4int\ ln(w)\ "d"w=1/4(wln(w)-int\ "d"w)
or simply $\frac{1}{4} w \ln \left(w\right) - \frac{1}{4} w + C$.

Substitute back $w = {x}^{4}$.

Our final answer is thus $\frac{1}{4} {x}^{4} \ln \left({x}^{4}\right) - \frac{1}{4} {x}^{4} + C = {x}^{4} \ln \left(x\right) - \frac{1}{4} {x}^{4} + C$.