What is the integral from 1 to t of 1/(x^2+4) ?

The integral from 1 to t of 1/(x^2+4) is 1/2 (arctan(t/2)-arctan(1/2))

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral from 1 to t of 1/(x^2+4)

\int_{1}^{t} \frac{1}{x ^{2} + 4} d x

\frac{1}{2} (arctan (\frac{t}{2}) - arctan (\frac{1}{2}))

Solution steps

\int_{1}^{t} \frac{1}{x ^{2} + 4} d x

Apply Integral Substitution

= \int_{\frac{1}{2}}^{\frac{t}{2}} \frac{1}{2 ( 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}{2} \cdot \int_{\frac{1}{2}}^{\frac{t}{2}} \frac{1}{u ^{2} + 1} d u

Use the common integral:

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

= \frac{1}{2} [arctan (u)]_{\frac{1}{2}}^{\frac{t}{2}}

Compute the boundaries:

arctan (\frac{t}{2}) - arctan (\frac{1}{2})

= \frac{1}{2} (arctan (\frac{t}{2}) - arctan (\frac{1}{2}))

\int_{- 6}^{- 2} 1 d x

\int_{\frac{π}{2}}^{π} cos^{2} (θ) d θ

\int_{0}^{2} 6 e^{- 2 t} d t

\int_{0}^{1} x^{3} - x^{4} d x

\int_{10}^{x} \frac{10}{x ^{2}} d x

What is the integral from 1 to t of 1/(x^2+4) ?
The integral from 1 to t of 1/(x^2+4) is 1/2 (arctan(t/2)-arctan(1/2))