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

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

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

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

3 π - \frac{3 π}{2}

Decimal Notation

4.71238 \dots

Solution steps

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

Compute the indefinite integral:

\int \frac{6}{x ^{2} + 1} d x = 6 arctan (x) + C

Compute the boundaries:

\int_{1}^{\infty} \frac{6}{x ^{2} + 1} d x = 3 π - \frac{3 π}{2}

= 3 π - \frac{3 π}{2}

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