What is the integral from 0 to t of 6/(1+6t^2) ?

The integral from 0 to t of 6/(1+6t^2) is sqrt(6)arctan(sqrt(6)t)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral from 0 to t of 6/(1+6t^2)

\int_{0}^{t} \frac{6}{1 + 6 t ^{2}} d t

6 arctan (6 t)

Solution steps

\int_{0}^{t} \frac{6}{1 + 6 t ^{2}} d t

Take the constant out:

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

= 6 \cdot \int_{0}^{t} \frac{1}{1 + 6 t ^{2}} d t

Apply Integral Substitution

= 6 \cdot \int_{0}^{6 t} \frac{1}{6 ( 1 + u ^{2} )} d u

Take the constant out:

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

= 6 \cdot \frac{1}{6} \cdot \int_{0}^{6 t} \frac{1}{1 + u ^{2}} d u

Use the common integral:

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

= 6 \cdot \frac{1}{6} [arctan (u)]_{0}^{6 t}

Simplify

6 \cdot \frac{1}{6} [arctan (u)]_{0}^{6 t} : 6 [arctan (u)]_{0}^{6 t}

= 6 [arctan (u)]_{0}^{6 t}

Compute the boundaries:

arctan (6 t)

= 6 arctan (6 t)

\int_{0}^{1} e^{- \frac{x ^{2}}{2}} d x

\int_{0}^{π} ∣ cos (x) - 0.5 ∣ d x

\int_{0}^{1} (ln (x))^{3} d x

\int_{0}^{5} x^{2} - 3 d x

\int_{1}^{5} 1 + x^{2} d x

What is the integral from 0 to t of 6/(1+6t^2) ?
The integral from 0 to t of 6/(1+6t^2) is sqrt(6)arctan(sqrt(6)t)