What is area y=(2x)/(1+x^2),-4<= x<= 4 ?

The answer to area y=(2x)/(1+x^2),-4<= x<= 4 is 2ln(17)

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area y=(2x)/(1+x^2),-4<= x<= 4

area

y = \frac{2 x}{1 + x ^{2}}, - 4 \leq x \leq 4

2 ln (17)

Decimal Notation

5.66642 \dots

Solution steps

Apply the area formula:

\int_{- 4}^{4} \frac{2 ∣ x ∣}{∣ 1 + x ^{2} ∣} d x

Solve

\int_{- 4}^{4} \frac{2 ∣ x ∣}{∣ 1 + x ^{2} ∣} d x : 2 ln (17)

The area is:

= 2 ln (17)

\frac{\partial}{\partial y} (\frac{y}{2 x})

\int_{1}^{2} (2 - x)^{- \frac{2}{3}} d x

y = \frac{x ^{6}}{4 - x ^{5}}

\frac{d}{d x} (ln (2 + x))

\int_{0}^{5} \frac{1}{x ^{\frac{12}{11}}} d x

What is area y=(2x)/(1+x^2),-4<= x<= 4 ?
The answer to area y=(2x)/(1+x^2),-4<= x<= 4 is 2ln(17)