What is the integral of 1/(\sqrt[4]{1-x^2)} ?

The integral of 1/(\sqrt[4]{1-x^2)} is 2\E((arcsin(x))/2 \vert 2)+C

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integral of 1/(\sqrt[4]{1-x^2)}

2 E (\frac{arcsin ( x )}{2} ∣2) + C

Solution steps

Apply Trigonometric Substitution

: \int cos (u) d u

= \int cos (u) d u

Use the nonelementary integral:

\int cos (x) d x = 2 E (\frac{x}{2} ∣ 2)

= 2 E (\frac{u}{2} ∣ 2)

Substitute back

u = arcsin (x)

= 2 E (\frac{arcsin ( x )}{2} ∣ 2)

Add a constant to the solution

= 2 E (\frac{arcsin ( x )}{2} ∣ 2) + C

\frac{x ^{2} + 9}{x + 3}

\int_{0}^{2} \frac{1}{4 + x ^{2}} d x

f (x) = \frac{2}{x + 1}, at x = 1

x \to 1 lim (x^{2} - x)

\frac{s ^{2} - 8 s + 15}{s ^{2} + 3 s + 2}

What is the integral of 1/(\sqrt[4]{1-x^2)} ?
The integral of 1/(\sqrt[4]{1-x^2)} is 2\E((arcsin(x))/2 \vert 2)+C