What is the integral from-1/3 to 1/3 of (1/(1+9x^2)) ?

The integral from-1/3 to 1/3 of (1/(1+9x^2)) is pi/6

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integral from-1/3 to 1/3 of (1/(1+9x^2))

\int_{- \frac{1}{3}}^{\frac{1}{3}} (\frac{1}{1 + 9 x ^{2}}) d x

\frac{π}{6}

Decimal

0.52359 \dots

Solution steps

\int_{- \frac{1}{3}}^{\frac{1}{3}} \frac{1}{1 + 9 x ^{2}} d x

Apply Integral Substitution

= \int_{- 1}^{1} \frac{1}{3 ( 1 + u ^{2} )} d u

Take the constant out:

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

= \frac{1}{3} \cdot \int_{- 1}^{1} \frac{1}{1 + u ^{2}} d u

Use the common integral:

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

= \frac{1}{3} [arctan (u)]_{- 1}^{1}

Compute the boundaries:

\frac{π}{2}

= \frac{1}{3} \cdot \frac{π}{2}

Simplify

= \frac{π}{6}

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What is the integral from-1/3 to 1/3 of (1/(1+9x^2)) ?
The integral from-1/3 to 1/3 of (1/(1+9x^2)) is pi/6