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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

= 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

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

\int 5 cos^{2} (x) d x

\frac{d}{d x} (- x^{3} + 9 x^{2} - 6 x + 9)

\frac{8}{64 + 64 t ^{2}}

\frac{d y}{d x} = 2 cos (x) (y + 1), y (\frac{π}{6}) = 2

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