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

The integral of (2x)/(x^2+y^4) is ln|y^4+x^2|+C

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integral of (2x)/(x^2+y^4)

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

ln y^{4} + x^{2} + C

Solution steps

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

Apply u-substitution

: \int \frac{1}{2 u} d u

= 2 \cdot \int \frac{1}{2 u} 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}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= 2 \cdot \frac{1}{2} ln ∣ u ∣

Substitute back

u = y^{4} + x^{2}

= 2 \cdot \frac{1}{2} ln y^{4} + x^{2}

Simplify

2 \cdot \frac{1}{2} ln y^{4} + x^{2} : ln y^{4} + x^{2}

= ln y^{4} + x^{2}

Add a constant to the solution

= ln y^{4} + x^{2} + C

f (x) = \frac{1}{2 x} x^{9}

(- 1, 7), (3, - 1)

x \to 0 lim (\frac{arccos ( 1 - x )}{x})

\int 1 - 2 x^{2} + x^{4} d x

x^{2} + y^{2} = 169, (- 12, 5)

What is the integral of (2x)/(x^2+y^4) ?
The integral of (2x)/(x^2+y^4) is ln|y^4+x^2|+C