What is the integral of (2x)/(sqrt(16x^4-9)) ?

The integral of (2x)/(sqrt(16x^4-9)) is 1/4 ln((|sqrt(16x^4-9)+4x^2|)/3)+C

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integral of (2x)/(sqrt(16x^4-9))

\int \frac{2 x}{16 x ^{4} - 9} d x

\frac{1}{4} ln (\frac{16 x ^{4} - 9 + 4 x ^{2}}{3}) + C

Solution steps

\int \frac{2 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

= 2 \cdot \int \frac{x}{16 x ^{4} - 9} d x

Apply u-substitution

: \int \frac{1}{2 16 u ^{2} - 9} d u

= 2 \cdot \int \frac{1}{2 16 u ^{2} - 9} 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}{16 u ^{2} - 9} d u

Apply Trigonometric Substitution

: \int \frac{sec ( v )}{4} d v

= 2 \cdot \frac{1}{2} \cdot \int \frac{sec ( v )}{4} d v

Take the constant out:

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

= 2 \cdot \frac{1}{2} \cdot \frac{1}{4} \cdot \int sec (v) d v

Use the common integral:

\int sec (v) d v = ln ∣ tan (v) + sec (v) ∣

= 2 \cdot \frac{1}{2} \cdot \frac{1}{4} ln ∣ tan (v) + sec (v) ∣

Substitute back

= 2 \cdot \frac{1}{2} \cdot \frac{1}{4} ln tan (arcsec (\frac{4}{3} x^{2})) + sec (arcsec (\frac{4}{3} x^{2}))

Simplify

2 \cdot \frac{1}{2} \cdot \frac{1}{4} ln tan (arcsec (\frac{4}{3} x^{2})) + sec (arcsec (\frac{4}{3} x^{2})) : \frac{1}{4} ln (\frac{16 x ^{4} - 9 + 4 x ^{2}}{3})

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

Add a constant to the solution

= \frac{1}{4} ln (\frac{16 x ^{4} - 9 + 4 x ^{2}}{3}) + C

2 x e^{x} cos (x)

\frac{d}{d x} (ln (\frac{x ^{\frac{1}{2}}}{3}))

\int (x^{2} - 3) d x

f (x) = ln (x^{3})

3 e^{3 t}

What is the integral of (2x)/(sqrt(16x^4-9)) ?
The integral of (2x)/(sqrt(16x^4-9)) is 1/4 ln((|sqrt(16x^4-9)+4x^2|)/3)+C