# How do you integrate int (z^2+1)/sqrt(z)dz?

Oct 10, 2015

I found: $2 \left({z}^{\frac{5}{2}} / 5 + \sqrt{z}\right) + c$

#### Explanation:

Let us set $z = {t}^{2}$ so:
$\mathrm{dz} = 2 t \mathrm{dt}$
and substituting:
$= \int \frac{{t}^{4} + 1}{\cancel{t}} \left(2 \cancel{t}\right) \mathrm{dt} =$
$2 \int \left({t}^{4} + 1\right) \mathrm{dt} = 2 \left({t}^{5} / 5 + t\right) + c =$
back to $z$:
$= 2 \left({z}^{\frac{5}{2}} / 5 + \sqrt{z}\right) + c$

Oct 10, 2015

$\frac{2}{5} {z}^{\frac{1}{2}} \left({z}^{\frac{4}{2}} + 5\right) + C \implies \frac{2}{5} \sqrt{z} \left({z}^{2} + 5\right) + C$

#### Explanation:

Separate and simplify the original expression into two parts then integrate each part. Sum the final integrations:

Write $\frac{{z}^{2} + 1}{\sqrt{z}} \text{ }$ as $\text{ } \frac{{z}^{2}}{{z}^{\frac{1}{2}}} + \frac{1}{{z}^{\frac{1}{2}}}$

Simplifying the indices giving:

${z}^{\frac{3}{2}} + {z}^{- \frac{1}{2}}$

Now integrate so we have:

$\int \left({z}^{\frac{3}{2}}\right) . \mathrm{dz} + \int \left({z}^{- \frac{1}{2}}\right) . \mathrm{dz}$

Giving:

$\left[\frac{{z}^{\frac{3}{2} + \frac{2}{2}}}{\frac{5}{2}} + {C}_{1}\right] + \left[\frac{{z}^{- \frac{1}{2} + \frac{2}{2}}}{\frac{1}{2}} + {C}_{2}\right]$

Let ${C}_{1} + {C}_{2} = {C}_{3}$ giving:

$\frac{2}{5} {x}^{\frac{5}{2}} + 2 {z}^{\frac{1}{2}} + {C}_{3}$

But $\frac{2}{5} {z}^{\frac{1}{2}} \times 5 = 2 {z}^{\frac{1}{2}}$

Factoring out $\frac{2}{5} {z}^{\frac{1}{2}}$ gives:

$\frac{2}{5} \sqrt{z} \left({z}^{2} + 5\right) + {C}_{3}$