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

The integral from 1 to 2 of 2/(9+4x^2) is 1/3 (arctan(4/3)-arctan(2/3))

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

\frac{1}{3} (arctan (\frac{4}{3}) - arctan (\frac{2}{3}))

Decimal Notation

0.11309 \dots

Solution steps

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

Take the constant out:

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

= 2 \cdot \int_{1}^{2} \frac{1}{9 + 4 x ^{2}} d x

Apply Integral Substitution

= 2 \cdot \int_{\frac{2}{3}}^{\frac{4}{3}} \frac{1}{6 ( u ^{2} + 1 )} d u

Take the constant out:

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

= 2 \cdot \frac{1}{6} \cdot \int_{\frac{2}{3}}^{\frac{4}{3}} \frac{1}{u ^{2} + 1} d u

Use the common integral:

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

= 2 \cdot \frac{1}{6} [arctan (u)]_{\frac{2}{3}}^{\frac{4}{3}}

Simplify

2 \cdot \frac{1}{6} [arctan (u)]_{\frac{2}{3}}^{\frac{4}{3}} : \frac{1}{3} [arctan (u)]_{\frac{2}{3}}^{\frac{4}{3}}

= \frac{1}{3} [arctan (u)]_{\frac{2}{3}}^{\frac{4}{3}}

Compute the boundaries:

arctan (\frac{4}{3}) - arctan (\frac{2}{3})

= \frac{1}{3} (arctan (\frac{4}{3}) - arctan (\frac{2}{3}))

\int_{1}^{3} 8 d x

\int_{0}^{\frac{π}{2}} cos (2 x - \frac{π}{2}) d x

\int_{0}^{1} r d r

\int_{0}^{4} 1 + 4 x^{2} d x

\int_{0}^{2} \frac{x}{( x ^{2} + 1 ) ^{2}} d x

What is the integral from 1 to 2 of 2/(9+4x^2) ?
The integral from 1 to 2 of 2/(9+4x^2) is 1/3 (arctan(4/3)-arctan(2/3))