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

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

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

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

arctan (x) - \frac{π}{4}

Solution steps

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

Use the common integral:

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

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

Compute the boundaries:

arctan (x) - \frac{π}{4}

= arctan (x) - \frac{π}{4}

\int_{0}^{\frac{π}{4}} sin^{2} (\frac{x}{2}) d x

\int_{0}^{\frac{π}{2}} 8 + 4 cos (x) d x

\int_{\frac{- π}{4}}^{\frac{π}{4}} cos (x) d x

\int_{0}^{\frac{π}{4}} tan^{3} (θ) d θ

\int_{0}^{2} 15 e^{3 x} d x

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