What is the integral from 0 to 1/3 of 5/(1+9x^2) ?

The integral from 0 to 1/3 of 5/(1+9x^2) is (5pi)/(12)

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integral from 0 to 1/3 of 5/(1+9x^2)

\int_{0}^{\frac{1}{3}} \frac{5}{1 + 9 x ^{2}} d x

\frac{5 π}{12}

Decimal Notation

1.30899 \dots

Solution steps

\int_{0}^{\frac{1}{3}} \frac{5}{1 + 9 x ^{2}} d x

Take the constant out:

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

= 5 \cdot \int_{0}^{\frac{1}{3}} \frac{1}{1 + 9 x ^{2}} d x

Apply Integral Substitution

= 5 \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

= 5 \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)

= 5 \cdot \frac{1}{3} [arctan (u)]_{0}^{1}

Simplify

5 \cdot \frac{1}{3} [arctan (u)]_{0}^{1} : \frac{5}{3} [arctan (u)]_{0}^{1}

= \frac{5}{3} [arctan (u)]_{0}^{1}

Compute the boundaries:

\frac{π}{4}

= \frac{5}{3} \cdot \frac{π}{4}

Simplify

= \frac{5 π}{12}

\frac{d}{d x} (\frac{1}{2} ln (\frac{x}{2}))

\frac{d}{d x} (x^{\frac{2}{3}} (x - 4))

\frac{d}{d x} ((2 x - 8)^{5})

f (t) = t^{2} e^{3 t}

\frac{d}{d x} (\frac{2 x + 3}{x + 2})

What is the integral from 0 to 1/3 of 5/(1+9x^2) ?
The integral from 0 to 1/3 of 5/(1+9x^2) is (5pi)/(12)