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

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

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

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

\frac{1}{3} (arctan (9) + arctan (6))

Decimal

0.95526 \dots

Solution steps

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

Apply Integral Substitution

= \int_{- 6}^{9} \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_{- 6}^{9} \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)]_{- 6}^{9}

Compute the boundaries:

arctan (9) + arctan (6)

= \frac{1}{3} (arctan (9) + arctan (6))

a re a x = 4 y - y^{2}, x = y^{2} - 7 y

\frac{\partial}{\partial y} (6 x^{2 y})

(x^{2} - 1) y^{^{'}} + 2 x y^{2} = 0, y (0) = 1

\frac{d}{d x} (sin (x^{2} + e^{x}))

\int_{0}^{π} 3 sin (x) d x

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