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

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

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integral from 0 to 1 of 6/(1+t^2)

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

\frac{3 π}{2}

Decimal Notation

4.71238 \dots

Solution steps

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

Take the constant out:

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

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

Use the common integral:

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

= 6 [arctan (t)]_{0}^{1}

Compute the boundaries:

\frac{π}{4}

= 6 \cdot \frac{π}{4}

Simplify

= \frac{3 π}{2}

\frac{d}{d x} (\frac{cos ( x )}{x} + \frac{x}{cos ( x )})

\frac{1}{4 x ^{2} + 1}

x \to 3 - lim (e^{\frac{3}{3 - x}})

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

\int 7^{x} d x

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