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

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

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

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

\frac{π}{2}

Decimal Notation

1.57079 \dots

Solution steps

\int_{1}^{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_{1}^{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)]_{1}^{3}

Compute the boundaries:

\frac{π}{12}

= 6 \cdot \frac{π}{12}

Simplify

= \frac{π}{2}

\int 6 e^{- 0.1 x} d x

sin (x), \frac{π}{2}

\int (x^{5} + x^{4})^{7} (5 x^{4} + 4 x^{3}) d x

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

x \to \infty lim (x (e^{\frac{1}{x}} - 1))

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