What is the integral from 0 to 5 of 1/(1+x^2) ?

The integral from 0 to 5 of 1/(1+x^2) is arctan(5)

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integral from 0 to 5 of 1/(1+x^2)

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

arctan (5)

Decimal Notation

1.37340 \dots

Solution steps

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

Use the common integral:

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

= [arctan (x)]_{0}^{5}

Compute the boundaries:

arctan (5)

= arctan (5)

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

x^{2} \frac{d y}{d x} - 2 x y = 6 y^{4}, y (1) = \frac{1}{4}

y^{^{''}} + y^{^{'}} = sec (x) tan (x)

f (x) = \frac{8 x}{8 + x}

What is the integral from 0 to 5 of 1/(1+x^2) ?
The integral from 0 to 5 of 1/(1+x^2) is arctan(5)