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

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

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

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

arctan (2) - \frac{π}{4}

Solution steps

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

Use the common integral:

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

= [arctan (x)]_{1}^{2}

Compute the boundaries:

arctan (2) - \frac{π}{4}

= arctan (2) - \frac{π}{4}

\int_{0}^{4} 2 x + 1 d x

\int_{0}^{\frac{π}{2}} 2 cos^{2} (θ) d θ

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

\int_{0}^{\frac{5}{8}} \frac{1}{64 x ^{2} + 25} d x

\int_{1}^{4} (x + \frac{1}{x})^{2} d x

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