Find the integral piint(81/(x^2+9))dx&-3&3

 Exercise

$\pi\int_{-3}^3\frac{81}{x^2+9}dx$

 

The integral of a function times a constant ($81$) is equal to the constant times the integral of the function

$254.4690049\int_{-3}^{3}\frac{1}{9+x^2}dx$

 

Solve the integral by applying the formula $\displaystyle\int\frac{x'}{x^2+a^2}dx=\frac{1}{a}\arctan\left(\frac{x}{a}\right)$

$\left[254.4690049\cdot \left(\frac{1}{3}\right)\arctan\left(\frac{x}{3}\right)\right]_{-3}^{3}$

 

Multiply the fraction and term in $254.4690049\cdot \left(\frac{1}{3}\right)\arctan\left(\frac{x}{3}\right)$

$\left[\frac{254.4690049}{3}\arctan\left(\frac{x}{3}\right)\right]_{-3}^{3}$

 

Evaluate the definite integral

$\frac{254.4690049}{3}\arctan\left(\frac{3}{3}\right)- \left(\frac{254.4690049}{3}\right)\arctan\left(-\frac{3}{3}\right)$

$\frac{254.4690049}{3}\arctan\left(\frac{3}{3}\right)- \left(\frac{254.4690049}{3}\right)\arctan\left(-\frac{3}{3}\right)$

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