What is the integral from 1 to infinity of 9/(1+x^2) ?

The integral from 1 to infinity of 9/(1+x^2) is (9pi)/4

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integral from 1 to infinity of 9/(1+x^2)

\int_{1}^{\infty} \frac{9}{1 + x ^{2}} d x

\frac{9 π}{4}

Decimal

7.06858 \dots

Solution steps

\int_{1}^{\infty} \frac{9}{1 + x ^{2}} d x

Compute the indefinite integral:

\int \frac{9}{1 + x ^{2}} d x = 9 arctan (x) + C

Compute the boundaries:

\int_{1}^{\infty} \frac{9}{1 + x ^{2}} d x = 9 \cdot \frac{π}{2} - 9 \cdot \frac{π}{4}

= 9 \cdot \frac{π}{2} - 9 \cdot \frac{π}{4}

Simplify

= \frac{9 π}{4}

\int_{0}^{2 π} 8 d t

\int_{1}^{4} x x - 1 d x

\int_{0}^{x} 1 - t^{2} d t

\int_{0}^{2} 2 x + x^{2} - x^{3} d x

\int_{0}^{5} ∣ 1 - x ∣ d x

What is the integral from 1 to infinity of 9/(1+x^2) ?
The integral from 1 to infinity of 9/(1+x^2) is (9pi)/4