What is the integral of 1/(xsqrt(64x^4-1)) ?

The integral of 1/(xsqrt(64x^4-1)) is 1/2 arctan(sqrt(64x^4-1))+C

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integral of 1/(xsqrt(64x^4-1))

\int \frac{1}{x 64 x ^{4} - 1} d x

\frac{1}{2} arctan (64 x^{4} - 1) + C

Solution steps

\int \frac{1}{x 64 x ^{4} - 1} d x

Apply u-substitution

: \int \frac{1}{4 u 64 u - 1} d u

= \int \frac{1}{4 u 64 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 64 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 (64 x^{4} - 1)

Simplify

\frac{1}{4} \cdot 2 arctan (64 x^{4} - 1) : \frac{1}{2} arctan (64 x^{4} - 1)

= \frac{1}{2} arctan (64 x^{4} - 1)

Add a constant to the solution

= \frac{1}{2} arctan (64 x^{4} - 1) + C

\frac{d}{d x} (\frac{1}{( x ^{3} - 2 x ^{2} + 7 ) ^{4}})

n = 1 \sum \infty 4 n - 1

\int \frac{1}{1 + 3 x ^{2}} d x

\frac{d}{d x} (\frac{3 x}{s q u ( x , q ) a r ( x , q ) e x ^{2} + 1})

What is the integral of 1/(xsqrt(64x^4-1)) ?
The integral of 1/(xsqrt(64x^4-1)) is 1/2 arctan(sqrt(64x^4-1))+C