What is the integral from 0 to 1 of (x/(x^2-5x+6))^2 ?

The integral from 0 to 1 of (x/(x^2-5x+6))^2 is -24ln(2)+12ln(3)+7/2

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral from 0 to 1 of (x/(x^2-5x+6))^2

\int_{0}^{1} (\frac{x}{x ^{2} - 5 x + 6})^{2} d x

- 24 ln (2) + 12 ln (3) + \frac{7}{2}

Decimal

0.04781 \dots

Solution steps

\int_{0}^{1} (\frac{x}{x ^{2} - 5 x + 6})^{2} d x

Apply exponent rule:

(\frac{a}{b})^{c} = \frac{a ^{c}}{b ^{c}}

= \int_{0}^{1} \frac{x ^{2}}{( x ^{2} - 5 x + 6 ) ^{2}} d x

Take the partial fraction of

\frac{x ^{2}}{( x ^{2} - 5 x + 6 ) ^{2}} : \frac{12}{x - 2} + \frac{4}{( x - 2 ) ^{2}} - \frac{12}{x - 3} + \frac{9}{( x - 3 ) ^{2}}

= \int_{0}^{1} \frac{12}{x - 2} + \frac{4}{( x - 2 ) ^{2}} - \frac{12}{x - 3} + \frac{9}{( x - 3 ) ^{2}} d x

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{12}{x - 2} d x + \int_{0}^{1} \frac{4}{( x - 2 ) ^{2}} d x - \int_{0}^{1} \frac{12}{x - 3} d x + \int_{0}^{1} \frac{9}{( x - 3 ) ^{2}} d x

\int_{0}^{1} \frac{12}{x - 2} d x = - 12 ln (2)

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

\int_{0}^{1} \frac{12}{x - 3} d x = 12 (ln (2) - ln (3))

\int_{0}^{1} \frac{9}{( x - 3 ) ^{2}} d x = \frac{3}{2}

= - 12 ln (2) + 2 - 12 (ln (2) - ln (3)) + \frac{3}{2}

Simplify

- 12 ln (2) + 2 - 12 (ln (2) - ln (3)) + \frac{3}{2} : - 24 ln (2) + 12 ln (3) + \frac{7}{2}

= - 24 ln (2) + 12 ln (3) + \frac{7}{2}

\int_{0}^{2 π} sin^{2} (2 θ) d θ

\int_{10}^{70} \frac{x - 10}{1800} d x

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

\int_{2}^{x} (\frac{1}{2} t^{2} - 1)^{8} d t

\int_{0}^{3} e^{1 - 6 t} d t

What is the integral from 0 to 1 of (x/(x^2-5x+6))^2 ?
The integral from 0 to 1 of (x/(x^2-5x+6))^2 is -24ln(2)+12ln(3)+7/2