What is the integral of (z^4)/(sqrt(1-z^{10))} ?

The integral of (z^4)/(sqrt(1-z^{10))} is 1/5 arcsin(z^5)+C

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integral of (z^4)/(sqrt(1-z^{10))}

\int \frac{z ^{4}}{1 - z ^{10}} d z

\frac{1}{5} arcsin (z^{5}) + C

Solution steps

\int \frac{z ^{4}}{1 - z ^{10}} d z

Apply u-substitution

: \int \frac{1}{5 1 - u ^{2}} d u

= \int \frac{1}{5 1 - u ^{2}} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{5} \cdot \int \frac{1}{1 - u ^{2}} d u

Use the common integral:

\int \frac{1}{1 - u ^{2}} d u = arcsin (u)

= \frac{1}{5} arcsin (u)

Substitute back

u = z^{5}

= \frac{1}{5} arcsin (z^{5})

Add a constant to the solution

= \frac{1}{5} arcsin (z^{5}) + C

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What is the integral of (z^4)/(sqrt(1-z^{10))} ?
The integral of (z^4)/(sqrt(1-z^{10))} is 1/5 arcsin(z^5)+C