What is the integral from 0 to 3/4 of 1/(9+16x^2) ?

The integral from 0 to 3/4 of 1/(9+16x^2) is pi/(48)

integral from 0 to 3/4 of 1/(9+16x^2)

\int_{0}^{\frac{3}{4}} \frac{1}{9 + 16 x ^{2}} d x

\frac{π}{48}

Decimal

0.06544 \dots

Solution steps

\int_{0}^{\frac{3}{4}} \frac{1}{9 + 16 x ^{2}} d x

Apply Integral Substitution

= \int_{0}^{1} \frac{1}{12 ( u ^{2} + 1 )} d u

Take the constant out:

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

= \frac{1}{12} \cdot \int_{0}^{1} \frac{1}{u ^{2} + 1} d u

Use the common integral:

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

= \frac{1}{12} [arctan (u)]_{0}^{1}

Compute the boundaries:

\frac{π}{4}

= \frac{1}{12} \cdot \frac{π}{4}

Simplify

= \frac{π}{48}

\frac{d}{d x} (\frac{1}{2 π} e^{- \frac{x ^{2}}{2}})

\frac{d}{d x} (A x^{2} e^{x} + B x e^{x} + C e^{x})

\int \frac{1}{10 x} d x

d er i v a t i v e y^{2}

f^{^{'}} (x) = 3 arctan (x^{2} - 3 x + 2)

What is the integral from 0 to 3/4 of 1/(9+16x^2) ?
The integral from 0 to 3/4 of 1/(9+16x^2) is pi/(48)