What is the integral from 0 to sqrt(3 of)(6/(1+x^2)) ?

The integral from 0 to sqrt(3 of)(6/(1+x^2)) is 2pi

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integral from 0 to sqrt(3 of)(6/(1+x^2))

\int_{0}^{3} (\frac{6}{1 + x ^{2}}) d x

2 π

Decimal Notation

6.28318 \dots

Solution steps

\int_{0}^{3} \frac{6}{1 + x ^{2}} d x

Take the constant out:

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

= 6 \cdot \int_{0}^{3} \frac{1}{1 + x ^{2}} d x

Use the common integral:

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

= 6 [arctan (x)]_{0}^{3}

Compute the boundaries:

\frac{π}{3}

= 6 \cdot \frac{π}{3}

Simplify

= 2 π

x \to 0 lim ((3 x - 4)^{4})

2 e^{\frac{s i n ( w )}{3}} + 1

\frac{d}{d x} (x^{2} 4 x - 3)

x \to 4 - lim (x^{2} + 2)

x \to 0 - lim (\frac{2}{9 x ^{\frac{1}{3}}})

What is the integral from 0 to sqrt(3 of)(6/(1+x^2)) ?
The integral from 0 to sqrt(3 of)(6/(1+x^2)) is 2pi