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

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

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

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

x arctan (\frac{y}{x}) + C

Solution steps

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

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

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

Apply Integral Substitution

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

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= x^{2} \frac{1}{x} \cdot \int \frac{1}{u ^{2} + 1} d u

Use the common integral:

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

= x^{2} \frac{1}{x} arctan (u)

Substitute back

u = \frac{y}{x}

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

Simplify

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

= x arctan (\frac{y}{x})

Add a constant to the solution

= x arctan (\frac{y}{x}) + C

\int x \frac{1}{1} d x

\frac{d}{d t} (b sin (wt))

\frac{d}{d x} (w \cdot sin (w c t) \cdot cos (w x))

\frac{\partial}{\partial x} (\frac{3}{x})

\frac{d x}{d t} = - 3 x (x^{2} - 1)

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