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

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

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integral of x/(x^2+y^2+1)

\int \frac{x}{x ^{2} + y ^{2} + 1} d x

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

Solution steps

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

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

Add a constant to the solution

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

f (x) = 2 x^{2} + 4 x + 3

\frac{d y}{d x} = \frac{2 y}{x}, x > 0

\frac{d}{d x} (\frac{x}{( 1 + 2 x ) ^{2}})

\frac{\partial}{\partial x} (ln (x + y^{3} + x y^{2}))

\frac{\partial}{\partial z} (x^{2} y z^{2})

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