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

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

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integral of 1/(xsqrt(x^4-4))

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

\frac{1}{4} arctan (\frac{1}{2} x^{4} - 4) + C

Solution steps

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

Apply u-substitution

: \int \frac{1}{4 ( u + 4 ) u} d u

= \int \frac{1}{4 ( u + 4 ) u} d u

Take the constant out:

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

= \frac{1}{4} \cdot \int \frac{1}{( u + 4 ) u} d u

Apply u-substitution

: \int \frac{2}{v ^{2} + 4} d v

= \frac{1}{4} \cdot \int \frac{2}{v ^{2} + 4} d v

Take the constant out:

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

= \frac{1}{4} \cdot 2 \cdot \int \frac{1}{v ^{2} + 4} d v

Apply Integral Substitution

= \frac{1}{4} \cdot 2 \cdot \int \frac{1}{2 ( w ^{2} + 1 )} d w

Take the constant out:

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

= \frac{1}{4} \cdot 2 \cdot \frac{1}{2} \cdot \int \frac{1}{w ^{2} + 1} d w

Use the common integral:

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

= \frac{1}{4} \cdot 2 \cdot \frac{1}{2} arctan (w)

Substitute back

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

Simplify

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

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

Add a constant to the solution

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

y^{^{''}} + 5 y = 4 t^{4}, y (0) = 0, y^{^{'}} (0) = 0

y = x^{4} - 17 x^{2} + 16

\frac{\partial}{\partial y} (x \cdot ln (x - y))

\int_{- 1}^{2} (2 x + 3) d x

x + 1

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