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

The integral of x/((x^2+y^2)) is 1/2 ln|y^2+x^2|+C

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

integral

\frac{x}{( x ^{2} + y ^{2} )}

\frac{1}{2} ln y^{2} + x^{2} + C

Solution steps

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

Apply u-substitution

= \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

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

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= \frac{1}{2} ln ∣ u ∣

Substitute back

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

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

Add a constant to the solution

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

in t e g r a l \frac{1}{( x ^{2} - a ^{2} )}

in t e g r a l ln (x^{2} + 1)

in t e g r a l \frac{1}{( a ^{2} + x ^{2} )}

in t e g r a l \frac{x}{( 1 - x ^{2} )}

in t e g r a l \frac{e ^{x}}{x ^{2}}

What is the integral of x/((x^2+y^2)) ?
The integral of x/((x^2+y^2)) is 1/2 ln|y^2+x^2|+C