What is the integral from 0 to 1/3 of 4/(1+9x^2) ?

The integral from 0 to 1/3 of 4/(1+9x^2) is pi/3

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integral from 0 to 1/3 of 4/(1+9x^2)

\int_{0}^{\frac{1}{3}} \frac{4}{1 + 9 x ^{2}} d x

\frac{π}{3}

Solution steps

\int_{0}^{\frac{1}{3}} \frac{4}{1 + 9 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_{0}^{\frac{1}{3}} \frac{1}{1 + 9 x ^{2}} d x

Apply Integral Substitution

= 4 \cdot \int_{0}^{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

= 4 \cdot \frac{1}{3} \cdot \int_{0}^{1} \frac{1}{1 + u ^{2}} d u

Use the common integral:

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

= 4 \cdot \frac{1}{3} [arctan (u)]_{0}^{1}

Simplify

4 \cdot \frac{1}{3} [arctan (u)]_{0}^{1} : \frac{4}{3} [arctan (u)]_{0}^{1}

= \frac{4}{3} [arctan (u)]_{0}^{1}

Compute the boundaries:

\frac{π}{4}

= \frac{4}{3} \cdot \frac{π}{4}

Simplify

= \frac{π}{3}

y = x^{3} - 2 x^{2} + 5

\int r ln (r) d r

x \to 1 lim (\frac{x ^{8} - 1}{x - 1})

n = 1 \sum \infty \frac{( 3 ^{n - 1} )}{9 ^{n}}

\frac{d}{d x} (4 x 16 - x^{2})

What is the integral from 0 to 1/3 of 4/(1+9x^2) ?
The integral from 0 to 1/3 of 4/(1+9x^2) is pi/3