What is the integral from 0 to infinity of 7/(x^2+1) ?

The integral from 0 to infinity of 7/(x^2+1) is (7pi)/2

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integral from 0 to infinity of 7/(x^2+1)

\int_{0}^{\infty} \frac{7}{x ^{2} + 1} d x

\frac{7 π}{2}

Decimal Notation

10.99557 \dots

Solution steps

\int_{0}^{\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_{0}^{\infty} \frac{7}{x ^{2} + 1} d x = 7 \cdot \frac{π}{2} - 0

= 7 \cdot \frac{π}{2} - 0

Simplify

= \frac{7 π}{2}

\int \frac{1}{x ^{8} + x ^{6}} d x

x \to 0 + lim (x (ln (4 x))^{2})

y = x^{4} + 5 x^{2} - x, (1, 5)

t d t + y e^{- t} d y = 0, y (0) = 1

y^{^{''}} + 4 y^{^{'}} + 3 y = e^{x}

What is the integral from 0 to infinity of 7/(x^2+1) ?
The integral from 0 to infinity of 7/(x^2+1) is (7pi)/2