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

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

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integral of (2x)/(sqrt(9+x^4))

\int \frac{2 x}{9 + x ^{4}} d x

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

Solution steps

\int \frac{2 x}{9 + x ^{4}} 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}{9 + x ^{4}} d x

Apply u-substitution

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

Apply Trigonometric Substitution

= 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) ∣

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

Substitute back

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

Simplify

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

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

Add a constant to the solution

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

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