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))

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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))

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))

x^{2} - 2 (- 4.14)

\int - sin (x) d x

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

\frac{1}{x + 1}, 1, 2

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))