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

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

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

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

\frac{3 π}{2}

Decimal

4.71238 \dots

Solution steps

\int_{0}^{1} \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}^{1} \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}^{1}

Compute the boundaries:

\frac{π}{4}

= 6 \cdot \frac{π}{4}

Simplify

= \frac{3 π}{2}

\frac{d}{d x} (\frac{π ( 4 x ^{2} + 4 x + 1 )}{16 x})

f (x) = \frac{1}{x ^{3}}, (2, \frac{1}{8})

\frac{d}{d x} (\frac{1}{x ^{2} + y ^{2} + z ^{2}})

\int e^{- x} cos (9 x) d x

\frac{d}{d x} (x (5 x + 9)^{\frac{1}{3}})

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