MHB How Do You Integrate (z^5 + 9z^2) * (z^3 + 1)^12?

[shamieh]

$$\int (z^5 + 9z^2) * (z^3 + 1)^{12} \, dz$$

$$u = z^3 + 1$$
$$z^3 = u - 1$$

$$du = 3z^2 $$

$$1/3 = z^2 dz$$

$$\int (z^5 + 9z^2) * (u - 1 + 1)^{12} $$

$$= \int (z^5 + 9z^2) * u^{12}$$

now I'm stuck..

 

[magneto1]

Re: taking A.D

Once you have $u = z^3 + 1$, and that $du = 3z^2 dz$.
Observe:
\[
(z^5 + 9z^2)dz = z^2(z^3 + 9)dz \Rightarrow \frac 13 (u-1+9) du...
\]

 

[shamieh]

Re: taking A.D

Once you have $u = z^3 + 1$, and that $du = 3z^2 dz$.
Observe:
\[
(z^5 + 9z^2)dz = z^2(z^3 + 9)dz \Rightarrow \frac 13 (u-1+9) du...
\]

Yea..Somehow my answer is incorrect?

Here is what I'm getting.

$$\frac{1}{3} \int z^2(z^3 + 9) * (u - 1 + 1)^{12} = \frac{1}{3} \int \frac{1}{3}(u - 1 + 9) * u^{12})$$

$$= \frac{1}{9} \int (u + 8) * u^{12} = \frac{1}{9} \int u^{13} + 8u^{12} = 1/9 ( \frac{u^{14}}{14} + \frac{8}{13} u^{13}) = \frac{(z^3 + 1)^{14}}{126} + \frac{8}{117} (z^3 + 1)^{13} + C$$

 

[magneto1]

Re: taking A.D

Is it me or is there an extra $\frac 13$ in the equation?

 

[Prove It]

Re: taking A.D

No the 1/3 is definitely needed...

$\displaystyle \begin{align*} \int{ \left( z^5 + 9z^2 \right) \, \left( z^3 + 1 \right) ^{12}\,\mathrm{d}z} &= \int{ z^2\,\left( z^3 + 9 \right) \, \left( z^3 + 1 \right) ^{12}\,\mathrm{d}z} \\ &= \frac{1}{3} \int{ 3z^2 \,\left( z^3 + 1 + 8 \right) \, \left( z^3 + 1 \right) ^{12}\,\mathrm{d}z } \end{align*}$

So now make the substitution $\displaystyle \begin{align*} u = z^3 + 1 \implies \mathrm{d}u = 3z^2\,\mathrm{d}z \end{align*}$ and the integral becomes

$\displaystyle \begin{align*} \frac{1}{3} \int{ 3z^2 \, \left( z^3 + 1 + 8 \right) \, \left( z^3 + 1 \right) \, \mathrm{d}z } &= \frac{1}{3} \int{ \left( u + 8 \right) \, u^{12} \, \mathrm{d}u } \\ &= \frac{1}{3} \int{ u^{13} + 8u^{12}\,\mathrm{d}u } \\ &= \frac{1}{3} \left( \frac{1}{14} u^{14} + \frac{8}{13} u^{13} \right) + C \\ &= \frac{1}{3} \left[ \frac{1}{14} \left( z^3 + 1 \right) ^{14} + \frac{8}{13} \left( z^3 + 1 \right) ^{13} \right] + C \end{align*}$