What is the integral from-1 to 1 of (x+3)/(x^2+2x+5) ?

The integral from-1 to 1 of (x+3)/(x^2+2x+5) is 1/2 ln(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-1 to 1 of (x+3)/(x^2+2x+5)

\int_{- 1}^{1} \frac{x + 3}{x ^{2} + 2 x + 5} d x

\frac{1}{2} ln (2) + \frac{π}{4}

Decimal

1.13197 \dots

Solution steps

\int_{- 1}^{1} \frac{x + 3}{x ^{2} + 2 x + 5} d x

Complete the square

x^{2} + 2 x + 5 : (x + 1)^{2} + 4

= \int_{- 1}^{1} \frac{x + 3}{( x + 1 ) ^{2} + 4} d x

Apply u-substitution

: \int_{0}^{2} \frac{u + 2}{u ^{2} + 4} d u

= \int_{0}^{2} \frac{u + 2}{u ^{2} + 4} d u

Apply Integral Substitution

= \int_{0}^{1} \frac{v + 1}{v ^{2} + 1} d v

Expand

\frac{v + 1}{v ^{2} + 1} : \frac{v}{v ^{2} + 1} + \frac{1}{v ^{2} + 1}

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int_{0}^{1} \frac{v}{v ^{2} + 1} d v + \int_{0}^{1} \frac{1}{v ^{2} + 1} d v

\int_{0}^{1} \frac{v}{v ^{2} + 1} d v = \frac{1}{2} ln (2)

\int_{0}^{1} \frac{1}{v ^{2} + 1} d v = \frac{π}{4}

= \frac{1}{2} ln (2) + \frac{π}{4}

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

\int_{0}^{t} e^{- u} (t - u)^{2} d u

\int_{0}^{\frac{π}{4}} (7 sec^{2} (x)) d x

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

\int_{0}^{5} 1 + t^{2} d t

What is the integral from-1 to 1 of (x+3)/(x^2+2x+5) ?
The integral from-1 to 1 of (x+3)/(x^2+2x+5) is 1/2 ln(2)+pi/4