What is the integral of (sqrt(z)+1/(3z))^2 ?

The integral of (sqrt(z)+1/(3z))^2 is -1/(9z)+4/3 sqrt(z)+(z^2)/2+C

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integral of (sqrt(z)+1/(3z))^2

\int (z + \frac{1}{3 z})^{2} d z

- \frac{1}{9 z} + \frac{4}{3} z + \frac{z ^{2}}{2} + C

Solution steps

\int (z + \frac{1}{3 z})^{2} d z

Expand

(z + \frac{1}{3 z})^{2} : \frac{1}{9 z ^{2}} + \frac{2}{3 z ^{\frac{1}{2}}} + z

= \int \frac{1}{9 z ^{2}} + \frac{2}{3 z ^{\frac{1}{2}}} + z d z

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int \frac{1}{9 z ^{2}} d z + \int \frac{2}{3 z ^{\frac{1}{2}}} d z + \int z d z

\int \frac{1}{9 z ^{2}} d z = - \frac{1}{9 z}

\int \frac{2}{3 z ^{\frac{1}{2}}} d z = \frac{4}{3} z^{\frac{1}{2}}

\int z d z = \frac{z ^{2}}{2}

= - \frac{1}{9 z} + \frac{4}{3} z^{\frac{1}{2}} + \frac{z ^{2}}{2}

Apply radical rule:

a^{\frac{1}{n}} = n a

= - \frac{1}{9 z} + \frac{4}{3} z + \frac{z ^{2}}{2}

Add a constant to the solution

= - \frac{1}{9 z} + \frac{4}{3} z + \frac{z ^{2}}{2} + C

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What is the integral of (sqrt(z)+1/(3z))^2 ?
The integral of (sqrt(z)+1/(3z))^2 is -1/(9z)+4/3 sqrt(z)+(z^2)/2+C