What is the integral from 0 to 6 of 1/(1+x^2) ?

The integral from 0 to 6 of 1/(1+x^2) is arctan(6)

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integral from 0 to 6 of 1/(1+x^2)

\int_{0}^{6} \frac{1}{1 + x ^{2}} d x

arctan (6)

Decimal Notation

1.40564 \dots

Solution steps

\int_{0}^{6} \frac{1}{1 + x ^{2}} d x

Use the common integral:

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

= [arctan (x)]_{0}^{6}

Compute the boundaries:

arctan (6)

= arctan (6)

\int_{0}^{\frac{π}{6}} (sec (x) + tan (x))^{2} d x

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\int_{0}^{1} \frac{x e ^{x}}{1 + e ^{x}} d x

\int_{0}^{\frac{π}{3}} tan (θ) d θ

What is the integral from 0 to 6 of 1/(1+x^2) ?
The integral from 0 to 6 of 1/(1+x^2) is arctan(6)