Problem 5: Evaluate $\displaystyle \int \dfrac{(6z - 1) \, dz}{\sqrt{(2z + 1)^3}}$ by Algebraic Substitution

$\displaystyle \int \dfrac{(6z - 1) \, dz}{\sqrt{(2z + 1)^3}}$

$\qquad = \displaystyle \int \dfrac{(6z - 1) \, dz}{\left[ (2z + 1)^{1/2} \right]^3}$
 

Let
$u = (2z + 1)^{1/2}$

$u^2 = 2z + 1$

$z = \frac{1}{2}(u^2 - 1)$

$dz = u \, du$

 

$\displaystyle \int \dfrac{(6z - 1) \, dz}{\sqrt{(2z + 1)^3}}$

$\qquad = \displaystyle \int \dfrac{\left[ 6 \cdot \frac{1}{2}(u^2 - 1) - 1 \right] (u \, du)}{u^3}$

$\qquad = \displaystyle \int \dfrac{(3u^2 - 4) \, du}{u^2}$

$\qquad = \displaystyle \int \left( 3 - \dfrac{4}{u^2} \right) \, du$

$\qquad = \displaystyle \int (3 - 4u^{-2}) \, du$

$\qquad = 3u + 4u^{-1} + C$

$\qquad = u^{-1} (3u^2 + 4) + C$

$\qquad = \dfrac{3u^2 + 4}{u} + C$
 

Revert to the original variable of integration
$\displaystyle \int \dfrac{(6z - 1) \, dz}{\sqrt{(2z + 1)^3}}$

$\qquad = \dfrac{3(2z + 1) + 4}{(2z + 1)^{1/2}} + C$

$\qquad = \dfrac{6z + 7}{\sqrt{2z + 1}} + C$   ←   answer