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

The integral from 1 to 3 of x/(5-2x+x^2) is 1/2 ln(2)+pi/8

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

\frac{1}{2} ln (2) + \frac{π}{8}

Decimal Notation

0.73927 \dots

Solution steps

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

Complete the square

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

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

Apply u-substitution

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

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

Expand

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

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

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

\int_{0}^{2} \frac{1}{u ^{2} + 4} d u = \frac{π}{8}

= \frac{1}{2} ln (2) + \frac{π}{8}

\int_{0}^{π} tan (2 x) d x

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

\int_{1}^{e} ln^{5} (x) \frac{1}{x} d x

\int_{1}^{2} 2 x + 3 d x

\int_{0}^{\frac{π}{4}} cos^{3} (θ) d θ

What is the integral from 1 to 3 of x/(5-2x+x^2) ?
The integral from 1 to 3 of x/(5-2x+x^2) is 1/2 ln(2)+pi/8