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

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

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

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

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

Solution steps

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

Apply u-substitution

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

= \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

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

Apply Trigonometric Substitution

: \int sec (v) d v

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

Use the common integral:

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

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

Substitute back

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

Simplify

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

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

Add a constant to the solution

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

\frac{1 + 0.5 x}{1 - 0.5 x}

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

\frac{\partial}{\partial x} (\frac{1}{2} (x^{2} + 6 y^{2}))

f (x) = x (x^{2} - 4 x + 5)^{4}, at x = 2

\frac{1}{s ^{2} + 2 s - 3}

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