What is the integral of (4x)/(sqrt(4x^4+9)) ?

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

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integral of (4x)/(sqrt(4x^4+9))

\int \frac{4 x}{4 x ^{4} + 9} d x

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

Solution steps

\int \frac{4 x}{4 x ^{4} + 9} d x

Take the constant out:

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

= 4 \cdot \int \frac{x}{4 x ^{4} + 9} d x

Apply u-substitution

: \int \frac{1}{2 4 u ^{2} + 9} d u

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

Take the constant out:

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

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

Apply Trigonometric Substitution

: \int \frac{sec ( v )}{2} d v

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

Take the constant out:

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

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

Use the common integral:

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

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

Substitute back

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

Simplify

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

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

Add a constant to the solution

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

y = 3 x, y = 2 x - 3, y = 0

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