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

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

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

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

\frac{π}{4}

Decimal Notation

0.78539 \dots

Solution steps

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

Take the constant out:

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

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

Apply Integral Substitution

= 3 \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

= 3 \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)

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

Simplify

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

= [arctan (u)]_{0}^{1}

Compute the boundaries:

\frac{π}{4}

= \frac{π}{4}

\int_{- 5}^{1} ∣ x + 3 ∣ d x

\frac{\partial ^{2}}{\partial y ^{2}} (x^{2} + y^{2})

y^{^{'}} + 9 x^{- 1} y = x^{7}, y (1) = 8

\int (5 x - 3)^{2} (2 x - 1) d x

x \to 0 lim (x ln (1 + \frac{1}{x}))

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