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

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

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integral from 2 to 3 of 4/(1+x^2)

\int_{2}^{3} \frac{4}{1 + x ^{2}} d x

4 (arctan (3) - arctan (2))

Solution steps

\int_{2}^{3} \frac{4}{1 + x ^{2}} d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 4 \cdot \int_{2}^{3} \frac{1}{1 + x ^{2}} d x

Use the common integral:

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

= 4 [arctan (x)]_{2}^{3}

Compute the boundaries:

arctan (3) - arctan (2)

= 4 (arctan (3) - arctan (2))

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

\int_{- 1}^{1} 3 x^{2} + 4 d x

\int_{0}^{2} x - 3 y^{2} d x

\int_{- 6}^{- 3} (x - \frac{3}{x ^{2}}) d x

\int_{0}^{4} e^{- s t} d t

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