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

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

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integral of (2x)/(x^2sqrt(x^4-1))

\int \frac{2 x}{x ^{2} x ^{4} - 1} d x

arctan (x^{4} - 1) + C

Solution steps

\int \frac{2 x}{x ^{2} x ^{4} - 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{x}{x ^{2} x ^{4} - 1} d x

Cancel the common factor:

x

= 2 \cdot \int \frac{1}{x x ^{4} - 1} d x

Apply u-substitution

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

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

Take the constant out:

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

= 2 \cdot \frac{1}{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 \cdot \frac{1}{2} arctan (u)

Substitute back

u = x^{4} - 1

= 2 \cdot \frac{1}{2} arctan (x^{4} - 1)

Simplify

2 \cdot \frac{1}{2} arctan (x^{4} - 1) : arctan (x^{4} - 1)

= arctan (x^{4} - 1)

Add a constant to the solution

= arctan (x^{4} - 1) + C

\frac{x ^{2}}{x ^{2} + 4}

x \to 4 lim (\frac{2}{( x - 4 ) ^{4}})

\int (e^{3 x} + \frac{1}{x}) d x

x \to 0 lim (\frac{2 ^{3 x} - 1}{x})

\int sin^{2} (θ) d θ

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