What is area f(x)= 1/(7(1+x^2)),g(x)= 1/(14x^2) ?

The answer to area f(x)= 1/(7(1+x^2)),g(x)= 1/(14x^2) is 1/7 arctan(2)+(-pi-1)/(28)

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area f(x)= 1/(7(1+x^2)),g(x)= 1/(14x^2)

area

f (x) = \frac{1}{7 ( 1 + x ^{2} )}, g (x) = \frac{1}{14 x ^{2}}

\frac{1}{7} arctan (2) + \frac{- π - 1}{28}

Decimal Notation

0.01025 \dots

Solution steps

Apply the area formula:

\int_{1}^{2} \frac{1}{7 ( 1 + x ^{2} )} - \frac{1}{14 x ^{2}} d x

Solve

\int_{1}^{2} \frac{1}{7 ( 1 + x ^{2} )} - \frac{1}{14 x ^{2}} d x : \frac{1}{7} arctan (2) + \frac{- π - 1}{28}

The area is:

= \frac{1}{7} arctan (2) + \frac{- π - 1}{28}

\int \frac{x ^{2} + 2}{x ^{3} + 3 x ^{2} + 2 x} d x

\frac{\partial}{\partial x} (4 x^{3} + x^{2} y - 3 y^{3})

\int_{1}^{e} ln (17 x) d x

\frac{d y}{d t} = 5

\frac{d}{d x} (arccos (e^{4 x}))

What is area f(x)= 1/(7(1+x^2)),g(x)= 1/(14x^2) ?
The answer to area f(x)= 1/(7(1+x^2)),g(x)= 1/(14x^2) is 1/7 arctan(2)+(-pi-1)/(28)