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}

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}

f (x) = 3 cot^{4} (x), at x = \frac{π}{4}

\int \frac{3}{x ^{2} x ^{2} + 9} d x

\frac{d}{d x} (x^{e^{2}})

\int \frac{e ^{x}}{1 + 64 e ^{2 x}} d x

e^{3 x} \cdot ln (4 x)

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