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

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

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integral of x/(x^2+y)

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

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

Solution steps

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

Apply u-substitution

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

= \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 + x^{2}

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

Add a constant to the solution

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

\frac{d}{d t} (e^{\frac{t ^{2}}{2}})

\int \frac{4}{x ^{2} - 16} d x

\frac{d}{d x} (3 x^{4 x})

\frac{\partial}{\partial x} (x \cdot \frac{e ^{y}}{x + y})

n = 8 \sum \infty \frac{n}{n ln ( n )}

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