What is the integral of sqrt(4/(x^4-x^2)) ?

The integral of sqrt(4/(x^4-x^2)) is 2arctan(sqrt(x^2-1))+C

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integral of sqrt(4/(x^4-x^2))

\int \frac{4}{x ^{4} - x ^{2}} d x

2 arctan (x^{2} - 1) + C

Solution steps

\int \frac{4}{x ^{4} - x ^{2}} d x

Simplify

\frac{4}{x ^{4} - x ^{2}} : \frac{2}{x ^{2} x ^{2} - 1}

= \int \frac{2}{x ^{2} x ^{2} - 1} d x

Take the constant out:

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

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

x^{2} = (x),

assuming

x \geq 0

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

Apply u-substitution

: \int \frac{1}{u ^{2} + 1} d u

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

Use the common integral:

\int \frac{1}{u ^{2} + 1} d u = arctan (u)

= 2 arctan (u)

Substitute back

u = x^{2} - 1

= 2 arctan (x^{2} - 1)

Add a constant to the solution

= 2 arctan (x^{2} - 1) + C

\int \frac{1}{x ^{2} x ^{2} - 5} d x

y = x^{8 c o s (x)}

y = arccsc (e^{9 t})

\int - \frac{8}{x ^{2} + 16} d x

f (x) = 8 e^{- 0.5 x}, at x = 7

What is the integral of sqrt(4/(x^4-x^2)) ?
The integral of sqrt(4/(x^4-x^2)) is 2arctan(sqrt(x^2-1))+C