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

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

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

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

- \frac{1}{2 ( y + x ^{2} )} + C

Solution steps

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

Apply u-substitution

: \int \frac{1}{2 ( y + u ) ^{2}} d u

= \int \frac{1}{2 ( y + 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}{( y + u ) ^{2}} d u

Apply u-substitution

: \int \frac{1}{v ^{2}} d v

= \frac{1}{2} \cdot \int \frac{1}{v ^{2}} d v

Apply the Power Rule

: - \frac{1}{v}

= \frac{1}{2} (- \frac{1}{v})

Substitute back

= \frac{1}{2} (- \frac{1}{y + x ^{2}})

Simplify

\frac{1}{2} (- \frac{1}{y + x ^{2}}) : - \frac{1}{2 ( y + x ^{2} )}

= - \frac{1}{2 ( y + x ^{2} )}

Add a constant to the solution

= - \frac{1}{2 ( y + x ^{2} )} + C

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