What is the integral from 1 to infinity of 7/(x^2+1) ?

The integral from 1 to infinity of 7/(x^2+1) is (7pi)/4

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integral from 1 to infinity of 7/(x^2+1)

\int_{1}^{\infty} \frac{7}{x ^{2} + 1} d x

\frac{7 π}{4}

Decimal Notation

5.49778 \dots

Solution steps

\int_{1}^{\infty} \frac{7}{x ^{2} + 1} d x

Compute the indefinite integral:

\int \frac{7}{x ^{2} + 1} d x = 7 arctan (x) + C

Compute the boundaries:

\int_{1}^{\infty} \frac{7}{x ^{2} + 1} d x = 7 \cdot \frac{π}{2} - 7 \cdot \frac{π}{4}

= 7 \cdot \frac{π}{2} - 7 \cdot \frac{π}{4}

Simplify

= \frac{7 π}{4}

y = ((\frac{x ^{2} - 2}{2 x ^{2} + 1})^{2})

\int 6 d x

\frac{d}{d x} (∣ 2 x - 1 ∣)

\frac{\partial}{\partial x} (\frac{( x - y )}{x + y})

\frac{d}{d x} ((x - 1)^{- 2})

What is the integral from 1 to infinity of 7/(x^2+1) ?
The integral from 1 to infinity of 7/(x^2+1) is (7pi)/4