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

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

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

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

y^{2} arctan (x) + C

Solution steps

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

Take the constant out:

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

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

Use the common integral:

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

= y^{2} arctan (x)

Add a constant to the solution

= y^{2} arctan (x) + C

f (t) = 7 t + 3

\int \frac{x}{( x + 2 ) ( 3 x - 1 )} d x

\int_{1}^{4} \frac{1}{x - 1} d x

\int ln (1 - x) d x

\frac{d y}{d t} = 2 (t + 1) y, y (0) = 1

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