What is the integral of 3/(4xsqrt(16x^4-9)) ?

The integral of 3/(4xsqrt(16x^4-9)) is 1/8 arctan((sqrt(16x^4-9))/3)+C

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integral of 3/(4xsqrt(16x^4-9))

\int \frac{3}{4 x 16 x ^{4} - 9} d x

\frac{1}{8} arctan (\frac{16 x ^{4} - 9}{3}) + C

Solution steps

\int \frac{3}{4 x 16 x ^{4} - 9} d x

Take the constant out:

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

= \frac{3}{4} \cdot \int \frac{1}{x 16 x ^{4} - 9} d x

Apply u-substitution

: \int \frac{1}{4 u 16 u - 9} d u

= \frac{3}{4} \cdot \int \frac{1}{4 u 16 u - 9} d u

Take the constant out:

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

= \frac{3}{4} \cdot \frac{1}{4} \cdot \int \frac{1}{u 16 u - 9} d u

Apply u-substitution

: \int \frac{2}{v ^{2} + 9} d v

= \frac{3}{4} \cdot \frac{1}{4} \cdot \int \frac{2}{v ^{2} + 9} d v

Take the constant out:

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

= \frac{3}{4} \cdot \frac{1}{4} \cdot 2 \cdot \int \frac{1}{v ^{2} + 9} d v

Apply Integral Substitution

= \frac{3}{4} \cdot \frac{1}{4} \cdot 2 \cdot \int \frac{1}{3 ( w ^{2} + 1 )} d w

Take the constant out:

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

= \frac{3}{4} \cdot \frac{1}{4} \cdot 2 \cdot \frac{1}{3} \cdot \int \frac{1}{w ^{2} + 1} d w

Use the common integral:

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

= \frac{3}{4} \cdot \frac{1}{4} \cdot 2 \cdot \frac{1}{3} arctan (w)

Substitute back

= \frac{3}{4} \cdot \frac{1}{4} \cdot 2 \cdot \frac{1}{3} arctan (\frac{16 x ^{4} - 9}{3})

Simplify

\frac{3}{4} \cdot \frac{1}{4} \cdot 2 \cdot \frac{1}{3} arctan (\frac{16 x ^{4} - 9}{3}) : \frac{1}{8} arctan (\frac{16 x ^{4} - 9}{3})

= \frac{1}{8} arctan (\frac{16 x ^{4} - 9}{3})

Add a constant to the solution

= \frac{1}{8} arctan (\frac{16 x ^{4} - 9}{3}) + C

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What is the integral of 3/(4xsqrt(16x^4-9)) ?
The integral of 3/(4xsqrt(16x^4-9)) is 1/8 arctan((sqrt(16x^4-9))/3)+C