What is the integral from 0 to 2 of 1/(x^2+4x+8) ?

The integral from 0 to 2 of 1/(x^2+4x+8) is 1/2 (arctan(2)-pi/4)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral from 0 to 2 of 1/(x^2+4x+8)

\int_{0}^{2} \frac{1}{x ^{2} + 4 x + 8} d x

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

Solution steps

\int_{0}^{2} \frac{1}{x ^{2} + 4 x + 8} d x

Complete the square

x^{2} + 4 x + 8 : (x + 2)^{2} + 4

= \int_{0}^{2} \frac{1}{( x + 2 ) ^{2} + 4} d x

Apply u-substitution

: \int_{2}^{4} \frac{1}{u ^{2} + 4} d u

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

Apply Integral Substitution

= \int_{1}^{2} \frac{1}{2 ( v ^{2} + 1 )} d v

Take the constant out:

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

= \frac{1}{2} \cdot \int_{1}^{2} \frac{1}{v ^{2} + 1} d v

Use the common integral:

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

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

Compute the boundaries:

arctan (2) - \frac{π}{4}

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

\int_{0}^{1} \frac{6 x ^{2}}{1 - x ^{3}} d x

\int_{0}^{\frac{1}{2}} \frac{1}{1 + 4 y ^{2}} d y

\int_{1}^{\infty} \frac{4}{3 x - 1} d x

\int_{0}^{\frac{π}{3}} 2 cos (x) - 1 d x

\int_{1}^{4} - 10 x + 50 d x

What is the integral from 0 to 2 of 1/(x^2+4x+8) ?
The integral from 0 to 2 of 1/(x^2+4x+8) is 1/2 (arctan(2)-pi/4)