What is the integral from 1 to 3 of 1/(1+x^2) ?

The integral from 1 to 3 of 1/(1+x^2) is arctan(3)-pi/4

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integral from 1 to 3 of 1/(1+x^2)

\int_{1}^{3} \frac{1}{1 + x ^{2}} d x

arctan (3) - \frac{π}{4}

Decimal

0.46364 \dots

Solution steps

\int_{1}^{3} \frac{1}{1 + x ^{2}} d x

Use the common integral:

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

= [arctan (x)]_{1}^{3}

Compute the boundaries:

arctan (3) - \frac{π}{4}

= arctan (3) - \frac{π}{4}

\int_{0}^{2} 3 t 4 - t^{2} d t

\int_{0}^{\frac{π}{3}} (7 sec^{2} (x)) d x

\int_{2}^{3} 2 x + 5 d x

\int_{a}^{b} g (x) d x

\int_{125}^{1} x^{3} d x

What is the integral from 1 to 3 of 1/(1+x^2) ?
The integral from 1 to 3 of 1/(1+x^2) is arctan(3)-pi/4