Integration by parts - order of $u$ and $v$

Question

I have the following integral to find:

$$\int 12x^2(3+2x)^5 dx$$

Now, I am aware of the integration by parts property -

$$\int \ u \frac{dv}{dx} = uv - \int v\frac{du}{dx}$$

Now, my question is the following -

When I make $u = 12x^2$, I find a different answer to when I make $u = (3+2x)^5$.

In integration by parts, can I make the value of $u$ whatever I want?

Answer 1

It does not matter to the final result what function you pick as $u$. You just need to determine what substitution will let you to get the result faster.

Answer 2

You will get the same answer which ever way you do it, but setting $u=(3+2x)^5$ is going to take a long time.

But an easier method is just a direct substitution $u=3+2x$, much quicker than integration by parts.

Answer 3

For solving integrals like this, with two small power and large power, we must exchange parenthesis by substitution. For instance I want to solve the integral $$I=\int 12x^2(3+2x)^{50} dx$$ which second has power $50$. With substitution $3+2x=u$ and $2dx=du$, the integral will simplify to $$I=\int 12\left(\frac{u-3}{2}\right)^2u^{50} \dfrac{du}{2}= \dfrac32\int(u^2-6u+9)u^{50}du=\dfrac32\int(u^{52}-6u^{51}+9u^{50})du$$ I think this method may help in such cases.

Answer 4

In addition to the answer already given (namely that that it does not matter which function you pick as "$ u $", as long as it's differentiable), may I suggest to make the substitution $ u = 3 + 2x $. Then \begin{align*} \int 12x^2(3+2x)^5 \: \mathrm{d}x &= \int 12 \left( \frac{u-3}{2} \right)^2 u^5 \: \frac{\mathrm{d}u}{2} \\ &= \frac{3}{2} \int \left( u^7 - 6u^6 + 9u^5 \right) \: \mathrm{d}u \\ &= \frac{3}{2} \int u^7 \: \mathrm{d}u - 9 \int u^6 \: \mathrm{d}u + \frac{27}{2} \int u^5 \: \mathrm{d}u , \end{align*} which is easy to evaluate. Then resubstitute to obtain an answer in $ x $. Beware, though, of adjusting the limits of integration when changing variables.

Answer 5

The answers to an integration may appear different without actually being different. For example:

$\int(x+2)^3\,dx$ by substitution, $\frac{1}{4}(x+2)^4 + C$.

By expanding first,

$\int(x+2)^3\,dx = \int(x^3+6x^2+12x+8)\,dx = \frac{x^4}{4}+2x^3 + 6x^2+8x+C$

These answers appear different.. But if you expand the first, you will see that they differ by only a constant. So the values of $C$ in the two cases are different.