What is area y=x^2,y= 2/(x^2+1) ?

The answer to area y=x^2,y= 2/(x^2+1) is -2/3+pi

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area y=x^2,y= 2/(x^2+1)

area

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

- \frac{2}{3} + π

Decimal Notation

2.47492 \dots

Solution steps

Apply the area formula:

\int_{- 1}^{1} x^{2} - \frac{2}{x ^{2} + 1} d x

Solve

\int_{- 1}^{1} x^{2} - \frac{2}{x ^{2} + 1} d x : - \frac{2}{3} + π

The area is:

= - \frac{2}{3} + π

x \to 1 - lim (x^{3} - 7 + m)

f (x) = \frac{3}{2 x + 5}

y^{^{'}} = 2 y + sin (2 t)

\frac{\partial}{\partial x} (e^{- (x^{3} + y^{3})})

\frac{\partial}{\partial x} (x^{3} + y^{3} - 5 x y)

What is area y=x^2,y= 2/(x^2+1) ?
The answer to area y=x^2,y= 2/(x^2+1) is -2/3+pi