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

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

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

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

arctan (x y) + C

Solution steps

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

Apply Integral Substitution

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

Take the constant out:

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

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

Use the common integral:

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

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

Substitute back

u = x y

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

Simplify

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

= arctan (x y)

Add a constant to the solution

= arctan (x y) + C

\frac{\partial}{\partial x} (ln (a x y) + b^{x y})

a re a 220 + 11 x - 11 x^{2}, 11 x^{2} - 55 x

d er i v a t i v e x^{5} e^{x}

\int 2 \cdot (x - 6)^{3} d x

t an g e n t f (x) = 7 + 4 x^{2} - 2 x^{3}, at x = a

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