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

\int \frac{1}{x ^{2}} 2 - \frac{1}{x} d x

(19 - (15 - 2 x^{4})^{7})^{3}

(12 x^{2} + y - 1) d x - (6 y - x) d y = 0

x y^{^{'}} + 5 y = 0

f (x) = e^{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