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

**bikerboy2442** · January 23rd 2013, 07:29 PM

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.

**Prove It** · January 23rd 2013, 07:36 PM

Originally Posted by bikerboy2442

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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.

First, rewrite this as $\displaystyle \begin{align*} -\int{-(z + 2)\sqrt{1 - z}\,dz} \end{align*}$ and then substitute $\displaystyle \begin{align*} u = 1 - z \implies du = -dz \end{align*}$ . Also note that if $\displaystyle \begin{align*} u = 1 - z \end{align*}$ , then $\displaystyle \begin{align*} z + 2 = 3 - u \end{align*}$ . The integral becomes

$\displaystyle \begin{align*} -\int{-(z + 2)\sqrt{1 - z}\,dz} &= -\int{ \left(3 - u \right) \sqrt{u}\,du} \\ &= -\int{ \left(3 - u \right) u^{\frac{1}{2}}\,du} \\ &= -\int{3u^{\frac{1}{2}} - u^{\frac{3}{2}}\,du} \end{align*}$

You should be able to go from here. Note there is no need for Integration by Parts.

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