Integral of (z+2)Sqrt(1-z)dz

[bikerboy2442]

I need to integrate this using u-substitution, integration by parts, and this integral table. I'm having a difficult time getting started. U substitution doesn't seem viable as I can't identify the derivative of any U within the integral. I'm assuming I need to use integration by parts, but internal integration and derivation only seems to provide a more difficult integral. The integral table doesn't seem to have any forms that are relevant to this problem.

 

[daon2]

Let u be 1-z.

Then z+2 = 3-u. And du = -dz.

 

[soroban]

Hello, bikerboy2442!

\(\displaystyle \displaystyle\int(z+2)\sqrt{1-z}\,dz\)(fixed a typo)

Here's a tip . . .

If the expression under the radical is linear,
. . let \(\displaystyle u\) equal the entire radical.

Let \(\displaystyle u \,=\, \sqrt{1-z} \quad\Rightarrow\quad u^2 \,=\,1-z \quad\Rightarrow\quad z \,=\,1-u^2\)

Then: .\(\displaystyle dz \,=\,-2u\,du \quad\Rightarrow\quad z+2 \,=\,3-u^2\)

Substitute: .\(\displaystyle \displaystyle\int(3-u^2)\cdot u \cdot(-2u\,du) \;=\; 2\int(u^4-3u^2)\,du \)

. . . . . . . . . \(\displaystyle =\;2\left(\frac{1}{5}u^5 - u^3\right) + C \;=\;\frac{2}{5}u^3(u^2-5) + C\)

Back-substitute: .\(\displaystyle \frac{2}{5}\left(\sqrt{1-z}\;\!\right)^3\left([\sqrt{1-z}\;\!]^2 - 5\right) + C \)

. . . . . . . . . . . . \(\displaystyle =\;\frac{2}{5}(1-z)^{\frac{3}{2}}(1-z-5) + C \)

. . . . . . . . . . . . \(\displaystyle =\;\frac{2}{5}(1-z)^{\frac{3}{2}}(-z-4) + C\)

. . . . . . . . . . . . \(\displaystyle =\;-\frac{2}{5}(1-z)^{\frac{3}{2}}(z+4) + C\)