What is the integral from 1 to sqrt(7 of) 9/(1+x^2) ?

The integral from 1 to sqrt(7 of) 9/(1+x^2) is 9(arctan(sqrt(7))-pi/4)

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integral from 1 to sqrt(7 of) 9/(1+x^2)

\int_{1}^{7} \frac{9}{1 + x ^{2}} d x

9 (arctan (7) - \frac{π}{4})

Decimal Notation

3.81627 \dots

Solution steps

\int_{1}^{7} \frac{9}{1 + x ^{2}} d x

Take the constant out:

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

= 9 \cdot \int_{1}^{7} \frac{1}{1 + x ^{2}} d x

Use the common integral:

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

= 9 [arctan (x)]_{1}^{7}

Compute the boundaries:

arctan (7) - \frac{π}{4}

= 9 (arctan (7) - \frac{π}{4})

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What is the integral from 1 to sqrt(7 of) 9/(1+x^2) ?
The integral from 1 to sqrt(7 of) 9/(1+x^2) is 9(arctan(sqrt(7))-pi/4)