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

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

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

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

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

Solution steps

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

Apply u-substitution

: \int \frac{1}{4 u 4 u - 1} d u

= \int \frac{1}{4 u 4 u - 1} 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 - 1} d u

Apply u-substitution

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

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

Use the common integral:

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

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

Substitute back

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

Simplify

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

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

Add a constant to the solution

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

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